Structure Of Atom

(i) As we know that

Mass of one electron = 9.1× 10^-28 g

1 electron weighs 9.1× 10^-28 g

So, X electrons will weigh 1 g

X = [1] / [9.1× 10^-28]

= 1.098×10²⁷ electrons

Therefore 1.098×10²⁷ electrons together weigh 1g.

(ii) By Avogadro’s Law we know that

1 mole of electron = 6.023×10²³ electrons

Mass of one electron = 9.1× 10^-28g

Therefore, Mass of 1 mole of electrons = [6.023×10²³] × [9.1× 10^-28]

= 5.4809× 10^-4g

We also know that charge of one electron = 1.6× 10^-19 C

Therefore, Charge of 1 mole of electron is = [6.023×10²³] × [1.6× 10^-19]

= 96.368×10³ C

Therefore, mass of one mole of electron is 5.481× 10^-4 g and the charge is 96.37×10³ C.

(i) As we know in 1 molecule of methane, 1 carbon atom and 4 atoms of hydrogen is present.

In carbon there are six electrons and hydrogen consist of one electron each.

So total number of electrons in methane = 6 + 4

= 10 electrons

By Avogadro’s Law we know that

1 mole of methane contains 6.023×10²³ atoms.

So total number of electrons of in 1 mole methane = 10×6.023×10²³

= 6.023×10²⁴ electrons

So total number of electrons in 1 mole of methane is 6.023×10²⁴ electrons

(ii) Mass of one neutron = 1.675 × 10^–27 kg

1 mole of Carbon atom = 6.023×10²³ atoms

Number of neutrons in 1 carbon atom = 14-6= 8

So total number of neutrons in 14g of Carbon = 6.023×10²³×8 neutrons

So 7mg of Carbon will contain =

= [3.37288×10²²]/14

= 2.4092×10²¹ neutrons

So, Mass of 2.4092×10²¹ neutrons = [2.4092×10²¹] × [1.675 × 10^–24]

= 4.035×10^–3 g

So, in 7 mg of carbon total number of neutrons is 2.41×10²¹ and the total mass of the neutrons is 4.035× 10^–3 g

(iii) Molecular Mass of Ammonia = 17g

By Avogadro’s Law,

1 mole of Ammonia = 17g of Ammonia = 6.023×10²³ atoms

Total Number of Protons in Ammonia = 7 + 3= 10

So total number of protons in Ammonia = 6.023×10²⁴ protons

17g of Ammonia contain 6.023×10²⁴ protons

So, 34 mg of Ammonia will contain X number of protons

X =

X = [6.023×10²⁴] × [2×10^-3]

X = 1.2046×10²² protons

Mass of one proton = 1.6726 × 10^–24g

So, Mass of 1.2046×10²² protons = [1.6726 × 10^–24] × [1.2046×10²²]

= 20.148×10^-3 g

So, in 34 mg of ammonia total number of protons is 1.205×10²² and the total mass of the protons is 20.148×10^-3 g.

(iv) No, the answer will not vary with the change in temperature and pressure because the number of subatomic particles like protons, neutrons and electrons is fixed for each and every element and it does not vary with temperature and pressure.

Suppose for element X with following representation

_zX^A

Where A = Mass Number of Element and Z = Atomic Number of the element.

(i) ₆C¹³

Mass Number of Carbon, A = 13

Atomic Number of Carbon, Z = 6

Number of protons = Number of Electrons = Atomic Number =6

Number of neutrons = A – Z

= 13-6 = 7

(ii) ₈O¹⁶

Mass Number of Oxygen, A = 16

Atomic Number of Oxygen, Z = 8

Number of protons = Number of Electrons = Atomic Number =8

Number of neutrons = A – Z

= 16-8= 8

(iii) ₁₂Mg²⁴

Mass Number of Magnesium, A = 24

Atomic Number of Magnesium, Z = 12

Number of protons = Number of Electrons = Atomic Number =12

Number of neutrons = A – Z

= 24-12 = 12

(iv) ₂₆Fe⁵⁶

Mass Number of Iron, A = 56

Atomic Number of Iron, Z = 26

Number of protons = Number of Electrons = Atomic Number = 26

Number of neutrons = A – Z

= 56-26 = 30

(v) ₃₈Sr⁸⁸

Mass Number of Strontium, A = 56

Atomic Number of Strontium, Z = 26

Number of protons = Number of Electrons = Atomic Number = 38

Number of neutrons = A – Z

= 88-38= 50

Suppose for element X with following representation

_zX^A

Where A = Mass Number of Element and Z = Atomic Number of the element.

(i) The representation with given atomic number and mass number is as follows:

Element with Atomic Number 17 is Chlorine.

₁₇Cl³⁵

(ii) The representation with given atomic number and mass number is as follows:

Element with Atomic Number 92 is Uranium.

₉₂U²³³

(iii) The representation with given atomic number and mass number is as follows:

Element with Atomic Number 4 is Beryllium.

₄Be⁹

Given:

Wavelength of Yellow Light = 580 nm.

To Find Frequency of the light:

We know the following basic relation,

Speed of Light = [Frequency] × [Wavelength]

We know speed of light = 3×10⁸ m/s

Frequency, v = [3×10⁸] / [580 × 10^-9]

v = 5.17×10¹⁴ Hz

Wave Number is defined as the number of wavelengths which can be accommodated in one centimetre of length in the direction of propagation.

Wave Number for Yellow Light = [1] / [λ]

= [1] / [580 × 10^-9]

= 1.7 × 10⁶ m^-1

Therefore, the frequency of yellow light is 5.17×10¹⁴ Hz and the wave number of yellow light is 1.7 × 10⁶ m^-1

By Planck’s Quantum Theory we have the following relation:

Energy, E = h×v

Where

h = Planck’s constant = 6.626×10^-34 Js

v = frequency

Using the Planck’s relation, we will solve the numerical.

(i) Frequency, v =3×10¹⁵ Hz

By Planck’s relation we have,

Energy, E = h×v

= [6.626×10^-34] × [3×10¹⁵]

= 1.9878×10^-18 J

Therefore, the energy of the photon corresponding to light of frequency 3×10¹⁵ Hz is 1.988×10^-18 J.

(ii) We know the following basic relation,

Speed of Light = [Frequency] × [Wavelength]

We know speed of light = 3×10⁸ m/s

Frequency, v = [3×10⁸] / [0.5 × 10^-10]

v = 6×10¹⁸ Hz

By Planck’s relation we have,

Energy, E = h×v

= [6.626×10^-34] × [6×10¹⁸]

= 3.9756×10^-15 J

Therefore, the energy of the photon corresponding to light of frequency 3×10¹⁵ Hz is 3.98×10^-15 J.

Given:

Period, T = 2.0 × 10^–10 s

Frequency, v = [1] / T

= [1] / [2.0 × 10^–10]

= 5.0 × 10⁹ Hz

We know the following basic relation,

Speed of Light = [Frequency] × [Wavelength]

We know speed of light = 3×10⁸ m/s

Wavelength, λ = [3×10⁸] / [5.0 × 10⁹]

= 0.06 m

Wave Number = [1] / [λ]

= [1/0.06]

= 16. 667 m^-1

Therefore, the frequency of wave is 5.0 × 10⁹ Hz and its wavelength is 0.06 m and its wave number 16. 667 m^-1

Given:

Wavelength, λ = 4000 pm

Energy, E = 1 J

By Planck’s relation we have,

Energy, E = h×v

But we know v = [c] / [λ]

Where

c = Speed of Light

v= Frequency

λ = Wavelength

So, E = hc /λ

= [[6.626×10^-34] × [3×10⁸]] / [4000×10^-12]

= [1.9878×10^-25] / [4000×10^-12]

= 4.9695×10^-17 J

4.9695×10^-17 J is the Energy of 1 photon

So, 1 J is the energy of X number of photons

X = [1] / [4.9695×10^-17]

X = 2.012×10¹⁶ photons

So, the number of photons which provide 1 J of energy is 2.012×10¹⁶ photons.

Given:

Wavelength, λ = 4 × 10^–7 m

Work Function of metal = 2.13 eV

Finding Energy of Photon:

By Planck’s relation we have,

Energy, E = h×v

But we know v = [c] / [λ]

Where

c = Speed of Light

v= Frequency

λ = Wavelength

So, E = hc /λ

= [[6.626×10^-34] × [3×10⁸]] / [4×10^-7]

= [1.9878×10^-25] / [4×10^-7]

= 4.9695×10^-19 J

Therefore, the energy of the photon 4.9695×10^-19 J

Finding the Kinetic Energy of Emission:

Kinetic Energy = hv - hv_o

= [E – W] eV

Where

E = Energy of photon in eV

W = work function of metal in eV

= {[4.9695×10^-19]/ [1.6020 × 10^–19]} – 2.13

= 3.102 – 2.13

= 0.972 eV

Therefore, the kinetic energy of emission is 0.972 eV.

Finding Velocity of photoelectron:

We know the formula for kinetic energy which is given as follows:

1/2 mv² = kinetic Energy

Where

m = mass of electron

v = velocity of electron

v² = [2× 0.972× 1.6020 × 10^–19] / [9.1× 10^-31]

= [3.1143× 10^–19] / [9.1× 10^-31]

v² = 3.422× 10¹¹

Therefore v = √ [3.422× 10¹¹]

v = 5.849× 10⁵ m/s

Therefore, the velocity of photoelectron is 5.85× 10⁵ m/s

Given:

Wavelength, λ = 242 nm

Finding Energy of Photon:

By Planck’s relation we have,

Energy, E = h×v

But we know v = [c] / [λ]

Where

c = Speed of Light

v= Frequency

λ = Wavelength

So, E = hc /λ

= [[6.626×10^-34] × [3×10⁸]] / [242×10^-9]

= [1.9878×10^-25] / [242×10^-9]

= 8.214×10^-19 J/atom

To convert the energy from J/atom to kJ mol^–1 we carry out the following conversion process:

E = [8.214×10^-19× 6.023×10²³] / 1000

= [494.729× 10³] / 1000

= 494.729 kJ/mol

Therefore, the energy of the photon 494.73 kJ/mol

Given:

Power of the bulb, P = 25 W

Wavelength of light emitted, λ = 0.57μm

By Planck’s relation we have,

Energy, E = h×v

But we know v = [c] / [λ]

Where

c = Speed of Light

v= Frequency

λ = Wavelength

So, E = hc /λ

= [[6.626×10^-34] × [3×10⁸]] / [0.57×10^-6]

= [1.9878×10^-25] / [0.57×10^-6]

= 3.487×10^-19 J

Rate of Emission of Quanta, R = P / E

Where

P = Power of the bulb

E = Energy of photon

R = [25] / [3.487×10^-19]

= 7.169×10¹⁹ s^-1

Therefore, the rate of emission of quanta is 7.17×10¹⁹ s^-1.

Given:

Wavelength, λ = 6800 Å

To find threshold frequency, v_o:

We know the following basic relation,

Speed of Light = [Frequency] × [Wavelength]

We know speed of light = 3×10⁸ m/s

Frequency, v_o = [3×10⁸] / [6800 × 10^-10]

v_o = 4.412×10¹⁴ Hz

Work Function, W₀ = [6.626×10^-34] × [4.412×10¹⁴]

= 2.923×10^-19 J

Therefore, the threshold frequency is 4.41×10¹⁴ Hz and the work function is 2.92×10^-19 J.

By referring to Balmer Formula:

We know that

Wave Number

Here n₁ = 2 and n₂ = 4 and the value of R_h = 109678

Wave Number

= [109678×3] / 16

= 329034 / 16

= 20564.625 cm^-1

We know that Wave Number = [1] / [Wavelength, λ]

Therefore, Wavelength = [1 / 20564.625]

λ = 4.863× 10^-5 cm

Therefore, the wavelength of the light is 4.86× 10^-5 cm

The expression of energy is given by,

E_n = - [2.18× 10^-18] Z²/n²

Where,

Z = atomic number of the atom

N = principal quantum number

For ionization from n1 = 5 to n₂ = ∞,

Therefore ∆E = E2-E1 = -21.8× 10^-19× [1/n₂²-1/n₁²]

= 21.8× 10^-19 × [1/n₂²-1/n₁²]

=21.8× 10^-19× [1/5²-1/∞]

=8.72× 10^-20J

For ionization from 1^st orbit, n₁=1, n₂=∞

Therefore ∆E’ = 21.8× 10^-19× [1/1²- 1/∞]

= 21.8× 10^-19J

Now ∆E’/∆E = 21.8× 10^-19/8.72× 10^-20 = 25

Thus, the energy required to remove electron from 1^st orbit is 25 times than the required to electron from 5^th orbit.

Given:

Present State of Electron = 6

When an electron drops from nth shell to ground state, the number of lines produced is given by = {n× [n-1]} / 2

So, here in this case, n= 6

So, number of lines produced in this transition is = {6× [6-1]}/2

=15

So, the lines produce in the transition is shown as below:

(i) Energy of an electrons = - (2.18× 10^-18)/n²

Where n = principal quantum number

Now the Energy associated with the fifth orbit of hydrogen atom is

E₅ = -(2.18× 10^-18)/(5)² = -2.18× 10^-18/25

E₅ = -8.72 × 10^-20J

(ii) Radius of Bohr’s n^th for hydrogen atom is given by,

R_n = (0.0529nm) / n2

For,

N = 5

R₅ = (0.0529 nm) (5)²

R₅ = 1.3225 nm

By referring to Rydberg Formula:

We know that

Wave Number

For Balmer series, the values of n₁ = 2 and the value of R_h = 109678

Wave number is inversely proportional to the wavelength of light. So for light of long wavelength, wave number has to be minimum. For minimum condition the value of n₂ should be equal to 3.

Wave Number

= 15233.0556 cm^-1

Therefore, the wave number for longest wavelength is 15.233× 10³ cm^-1.

Given:

Ground State Electron energy = –2.18 × 10^–11 ergs

Finding Energy:

To convert energy from ergs to joules

1 ergs is equal to 10^-7 Joules

So –2.18 × 10^–11 ergs = –2.18 × 10^–11 × 10^-7

So, Ground State Electron energy = = –2.18 × 10^–18 J

Energy to shift the electron from n = 1 to n= 5 state is given by following relation:

Eh = E₅ – E₁

The energy of hydrogen atom is given by the following equation:

En = -2n²me⁴Z²/n²h²

Where

m = mass of electrons

Z = atomic mass of atom

E = charge of electron

h = Planck’s constant

So, the energy required is =

=

= 2.0928 × 10^–18 J

Therefore, the energy to shift the electron from n=1 to n=5 state is 2.093 × 10^–18 J.

Finding Wavelength:

By Planck’s relation we have,

Energy, E = h×v

But we know v = [c] / [λ]

Where

c = Speed of Light

v= Frequency

λ = Wavelength

So E = hc /λ

λ = hc / E

= [[6.626×10^-34] × [3×10⁸]] / [2.0928 × 10^–18]

= [1.9878×10^-25] / [2.0928 × 10^–18]

= 9.498×10^-8 m

Therefore, the wavelength is 9.5×10^-8 m

Given:

Electron energy in Hydrogen atom E_n = [(–2.18 × 10^–18)/n²] J

Energy for first state, E₁ = [–2.18 × 10^–18]/1²

= –2.18 × 10^–18 J

Energy for second state, E₂ = [–2.18 × 10^–18]/2²

= –0.5465 × 10^–18 J

By Planck’s relation we have,

Energy, E = h×v

But we know v = [c] / [λ]

Where

c = Speed of Light

v= Frequency

λ = Wavelength

So, E = hc /λ

λ = hc / E

= [[6.626×10^-34] × [3×10⁸]] / [0.5465 × 10^–18]

= [1.9878×10^-25] / [0.5465 × 10^–18]

= 3.637×10^-7 m

Therefore, the wavelength is 3.64×10^-7 m

Given:

Velocity of electron = 2.05 × 10⁷ m s^–1

According to the de Broglie’s Equation:

λ = h/mv

Where,

λ = wavelength of moving particle

m = mass of electron

v= velocity of particle

h= Planck’s constant [6.62× 10^-34]

λ = {[6.62× 10^-34] / [9.1× 10^-31] × [2.05 × 10⁷]}

= {[6.62× 10^-34] / [1.8655 × 10^-23]

= 3.552× 10^-11 m

Therefore, the wavelength is 3.55× 10^-11 m

Given:

Mass of an Electron = 9.1 × 10^–31 kg

Kinetic Energy, K.E. = 3.0 × 10^–25 J

1/2 mv² = kinetic Energy

Where

m = mass of electron

v = velocity of electron

v² = [2× 3.0 × 10^–25] / [9.1× 10^-31]

= [6.0 × 10^–25] / [9.1× 10^-31]

v² = 659.3406 × 10³

Therefore v = √ [659.3406 × 10³]

v = 811.99 m/s

According to the de Broglie’s Equation:

λ = h/mv

Where,

λ = wavelength of moving particle

m = mass of electron

v= velocity of particle

h= Planck’s constant [6.62× 10^-34]

λ = {[6.62× 10^-34] / [9.1× 10^-31] × [811.99]}

= {[6.62× 10^-34] / [7.389 × 10^-28]

= 8.959× 10^-7 m

Therefore, the wavelength is 8.96× 10^-7 m

Isoelectronic species are defined as those species which belong to different atoms or ions which possess same number of electrons but different magnitude of nuclear charge.

A positive charged ion denotes the loss of an electron & A negative charged ion represents the gain of an electron by a species.

1] Number of electrons in sodium [Na] = 11

Therefore, Number of electrons in sodium ion [Na⁺] = 10

2] Number of electrons in potassium ion [K⁺] = 18

3] Number of electrons in magnesium ion [Mg²⁺] = 10

4] Number of electrons in calcium ion [Ca²⁺] = 18

5] Number of electrons in sulphur [S] = 16

∴ Number of electrons in sulphur ion [S^2-] = 18

6] Number of electrons in argon [Ar] = 18

Hence, the following ions are isoelectronic species:

1] Na⁺ and Mg²⁺ [10 electrons each]

2] K⁺, Ca²⁺, S^2- and Ar [18 electrons each]

The electronic configuration of ions refers to representation of electrons in the various energy shells and subshells more collectively called as orbits of the atom.

[a] H^- ion

The electronic configuration of H atom is 1s¹. [Atomic number = 1]

∴ Electronic configuration of H^- = 1s²

[b] Na⁺ ion

The electronic configuration of Na atom is 1s² 2s² 2p⁶ 3s¹. [Atomic number = 11]

∴ Electronic configuration of Na⁺ = 1s₂ 2s² 2p⁶3s⁰ Or 1s² 2s² 2p⁶

[c] O^2- ion

The electronic configuration of 0 atom is 1s² 2s² 2p⁴. [Atomic number = 8]

∴ Electronic configuration of O^2- ion = 1s² 2s² p⁶

[d] F^- ion

The electronic configuration of F atom is 1s² 2s² 2p⁵. [Atomic number = 9]

∴ Electron configuration of F^- ion = 1s² 2s² 2p⁶

[a] 3s¹

Completing the electron configuration of the element as 1s² 2s² 2p⁶ 3s¹.

∴ Number of electrons present in the atom of the element

= 2 + 2 + 6 + 1 = 11

∴ Atomic number of the element = 11 [sodium]

[b] 2p³

Completing the electron configuration of the element as 1s² 2s² 2p3

∴ Number of electrons present in the atom of the element = 2+ 2 + 3 = 7

∴ Atomic number of the element = 7 [nitrogen]

[c] 3p⁵

Completing the electron configuration of the element as 1s² 2s² 2p⁶ 3s² 3p⁵

∴ Number of electrons present in the atom of the element = 2 + 2 + 6+2+5 =17

∴ Atomic number of the element = 17 [Chlorine]

[a] [He] 2s¹

The electronic configuration of Helium is [He] 2s1 = 1s² 2s¹.

∴ Atomic number of the element = 3 [lithium, a p-block element]

[b] [Ne] 3s² 3p³

The electronic configuration of Neon is [Ne] 3s² 3p³ = 1s² 2s² 2p⁶ 3s² 3p³.

∴ Atomic number of the element = 15 [phosphorous, a p block element]

[c] [Ar] 4s² 3d¹

The electronic configuration of Argon is given as [Ar] 4s² 3d¹ = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹.

∴ Atomic number of the element = 21 [scandium, a d block element]

Quantum number refers to the set of four special numbers which provide the entire information about electrons of a particular atom.

Thus, the letter ‘n’ represents the principle quantum number and the Azimuthal quantum number ‘l’ can have any value between to n-1.

For n = 1 [ K shell] has L = 0 [ one subshell]

For n = 2 [L shell] has L = 0, 1 [2 subshell]

For n = 3 [M shell] has L = 0, 1, 2 [3 subshell]

For n = 4 [N shell] has L = 0, 1, 2, 3, [4 subshell]

For n = 5 [O shell] has L = 0, 1, 2, 3, 4, [5 subshell]

∴ For l = 4, minimum value of n = 5

‘n’ refers to the principle quantum number

‘l’ refers to the azimuthal quantum number

‘m’ refers to the magnetic quantum number

When l = 2 the value of m = -2, -1, 0, +1, +2

Now, for the 3d orbital:

Principal quantum number [n] = 3

Azimuthal quantum number [L] = 2

Magnetic quantum number [m] = -2, -1, 0, 1, 2

(i) In an atom the number of electrons is always equal to the number of protons to balance the charge in the atom and thus an atom is electrically neutral.

So, number of protons = number of electrons = 29

(ii) The electronic configuration of the atom is given as below

1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d¹⁰

This electronic configuration refers to the electronic configuration of copper.

A positive charged ion denotes the loss of an electron & A negative charged ion represents the gain of an electron by a species.

For H⁺₂:

Number of electrons present in Hydrogen molecule = 2

So, Number of electrons present in H⁺₂ = 2 – 1

= 1

For H₂:

Number of electrons present in Hydrogen molecule = 2

For O⁺₂:

Number of electrons present in Oxygen molecule = 8 + 8

= 16

So, Number of electrons present in O⁺₂ = 16 - 1

= 15

(i) ‘n’ refers to the principle quantum number

‘l’ refers to the azimuthal quantum number

‘m’ refers to the magnetic quantum number

l can have any value from to n – 1.

So, for n= 3, the permissible value of l = 0, 1, 2

M can have any value from –l, -l+1 ………..0, 1 ……… l

For l = 0

m = 0

For l = 1

m = +1, 0, -1

For l = 2

m = +2, +1, 0, -1, -2

(ii) When l = 2 the value of m = -2, -1, 0, +1, +2

Now, for the 3d orbital:

Principal quantum number [n] = 3

Azimuthal quantum number [L] = 2

Magnetic quantum number [mL] = -2, -1, 0, 1, 2

(iii) For a given value of n the value of l can range from 0 to n-1
So for n =1, l = 0

Thus, 1p is not possible
For n = 2, l = 0 and 1

Therefore, 2s and 2p are possible orbitals.
For n = 3, l = 0, 1 and 2

So 3p is not possible.

‘n’ refers to the principle quantum number

‘l’ refers to the azimuthal quantum number

‘m’ refers to the magnetic quantum number

[a] when n = 1, L = 0 [Given] The orbital is 1s. [Can have maximum of 2 electrons]

[b] For n = 3 and L = 1 the orbital is 3p. [Can have maximum of 6 electrons].

[c] For n = 4 and L = 2 the orbital is 4d. [Can have maximum of 10 electrons]

[d] For n = 4 and L = 3, the orbital is 4f. [Can have maximum of 14 electrons]

we know
N : It is the principal quantum numbers and it can take any value from 1 to n(arbitrary number)
L : It is the number of the sub-shells of the orbit can take any value from 0 to (n-1)
specifies the orientation of the orbit (ml=2×L+1) and it can take any value from -L to L
m_s : It specifies spin of the atom and it can take values ms = -1/2 and 1/2

now, let's solve the options:-

(a) The given set of quantum numbers is wrong because the value of principal quantum number that is [n] cannot be zero_.

(b) This set of quantum numbers is possible as n=1>0,L=(n-1)= 0 is possible,ml=0 is also possible as l=0, and ms=-1/2 is also possible.

(c) This set is not possible because for principle quantum number n = 1, l=(n-1)=0 but L=1 according to question.

(d) This set of quantum numbers is possible as n=2>0,L=2-1=1,no. of ml=2×1+1=3 which are -1,0,1,ms=-1/2 is also possible .

(e) This set is not possible because for principle quantum number n = 3, the quantum number ‘l’ =3-1=2<3(given in question).

(f) This set of quantum numbers is possible as n=3>0,l=3-1=2 can take any value from the value 0,1,2 and ml=2×l+1=-1,0,1 and ms=+1/2 is also possible .

(a) Total number of electrons in an atom can be found out by the following relationship

n_e = 2n²

Total no of electrons when n=4, then n = 2× 4² = 32 & half of them have m_s = -1/2

(b) When n = 3, L=0 means 3s orbital which can have 2 electrons.

As we know that the angular momentum of an electron is given by mvr = [nh]/ 2π ……………………………………………. [1]

Where

m = mass of the electron

v = velocity of the electron

r = radius of the orbit

h = Planck’s constant

According to the de Broglie’s Equation:

λ = h/mv

Where,

λ = wavelength of moving particle

m = mass of electron

v= velocity of particle

h= Planck’s constant [6.62× 10^-34]

mv = h / λ

Substituting the above value in [1]

[hr] / λ = [nh]/ 2π

2πr = nλ

Since we know that 2πr is the circumference of a circle, so it is proved in above equation that circumference of the Bohr’s orbit is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit.

For an atom, wave number = 1/λ = R_HZ² [1/n₁²-1/n₂²]

For He⁺ spectrum Z = 2, n₂=4, n₁=2

Therefore [^-v] = 1/λ = R_H× {4× [1/2²-1/4²]}

= R_H× {4× [1/4 – 1/16]}

= R_H× {[4× 3] / 16}

= 3R_H / 4

For hydrogen spectrum, wave number = 3R_H/4, Z = 1

Therefore, wave number = 1/λ = R_H× 1[1/n₁²-1/n₂²]

Or

R_H [1/n₁²-1/n₂²] = 3R_H/4

Or

1/n₁² – 1/n₂² = 3/4

Which can be so for n₁ = 1 & n₂ = 2, i.e. the transition is from n = 2 to n = 1

Given:

Ground State Energy = 2.18 × 10^–18 J atom^–1

The energy of hydrogen atom is given by the following equation:

En = -2n²me⁴Z²/n²h²

Where

m = mass of electrons

Z = atomic mass of atom

E = charge of electron

h = Planck’s constant

For the process mentioned in the question, energy required = E_n– E₁

= 0 – [-2n²me⁴ / 1²h²]

= 4 × 2n²me⁴/h²

But we know 2n²me⁴/h² = 2.18 × 10^-18J

= 4 × 2.18 × 10^-18

= 8.72 × 10^-18J

Therefore, the energy required for the process is 8.72 × 10^-18J

Given:

Diameter of carbon atom = 0.15 nm

Radius of carbon atom = [0.15 / 2] = 0.075 nm

Length of line = 20 cm

Number of carbon atoms which can be place along the line is given by n = [20× 10^-2] / [0.15× 10^-9]

n = 1.333× 10⁹

Therefore, the number of carbon atoms which can be place along the line is 1.333× 10⁹.

Given:

Number of carbon atoms, n = 2 ×10⁸

Length of the arrangement, l = 2.4 cm

Number of carbon atoms which can be place along the line is given by n = [l] / [diameter, d]

D= [l] / [n]

D = [2.4× 10^-2] / [2 ×10⁸]

D = 1.2× 10^-10 m

Therefore radius = [1.2× 10^-10] / 2

= 0.6× 10^-10 m

Therefore, the radius of carbon atom is 0.6 Å.

Given:

Diameter of carbon atom = 2.6 Å

Radius of carbon atom = [2.6 / 2] = 1.3 Å

Length of line = 1.6 cm

Number of carbon atoms which can be place along the line is given by n = [1.6× 10^-2] / [2.6× 10^-10]

n = 61.538× 10⁶

Therefore, the number of carbon atoms which can be place along the line is 61.538× 10⁶.

Given:

Charge on particle = 2.5 × 10^–16C

Charge on electron, = ne

Where

n = number of electrons

e = 1.6× 10^-19 J

n = [2.5 × 10^–16] / [1.6× 10^-19]

n = 1562 electrons

Therefore, the number of electrons present is 1562 electrons.

Given:

Charge present on oil drop = –1.282 × 10^–18 C

Charge of one electron = 1.6× 10^-19 J

Charge on oil drop, = ne

Where

n = number of electrons

e = 1.6× 10^-19 J

n = [–1.282 × 10^–18 C] / [1.6× 10^-19]

n = 8 electrons

Therefore, the number of electrons present on oil drop is 8 electrons.

In 1911, Rutherford performed alpha rays scattering experiment to demonstrate and to find the structure of atom.

Heavy atoms have a heavy nucleus carrying a great quantity of positive charge. Hence, some alpha particles are easily deflected back on hitting the nucleus. Besides a bit of alpha particles are deflected through small angles because of large positive charge on the nucleus. So with the aid of heavy nuclei atoms it was possible to understand or detect the minute deflections of alpha particles.

If light atoms are used, their nuclei will be light & moreover, they will hold a small positive charge on the nucleus. Hence, the number of particles deflected back & those deflected through some angle will be trifling. Hence deflections of alpha particles from light atoms remain unnoticeable.

The general convention or method of representing an element along with its atomic mass [A] and atomic number [Z] followed globally is ^A_ZX. Atomic number of an element is fixed. However, mass number is not fixed as it depends upon the isotope taken. Hence it is essential to indicate mass number.

Given:

Mass Number = 81

Percentage of neutron = 31.7% more as compared protons

Let number of protons = p

Let number of neutrons= n

p + n = 81

p + {p + [31.7× p] / 100} = 81

p + 1.317p = 81

2.317p = 81

p = 81 / 2.317

p = 94.95899

p = 35

Atomic Number = 35

The element is bromine.

The atomic symbol is 81₃₅Br.

Given:

Mass Number = 37

Percentage of neutron = 11.1% more as compared electrons

Let Number of electrons = number of protons = p

Let number of protons = p

Let number of neutrons= n

p + n = 81

p + {p + [11.1× p] / 100} = 37

p + 1.111p = 37

2.111p = 37

p = 37 / 2.111

p = 17.527

p = 17

Atomic Number = 17

The element is chlorine.

The atomic symbol is 37₁₇Cl.

Given:

Mass Number = 56

Percentage of neutron = 30.4% more as compared electrons

Let Number of electrons in ion = p

Therefore, number of protons = p + 3

Therefore, number of neutrons= p + 0.304 p =1.304 p

As, we all know that mass number is the sum of number of protons and neutrons.

Therefore,

p + 3 + n = 56

p + 3 + {p + [30.4× p] / 100} = 56

p + 3 + 1.304p = 56

2.304p = 53

p = 53 / 2.304

p = 23

Therefore, the number of protons = p + 3

=23 + 3

= 26

Atomic Number = 26

The element is Iron.

The atomic symbol is Fe³⁺.

As the frequency increases the wavelength decreases. The arrangement is given as below:

Cosmic rays < X-rays < radiation from microwave ovens < amber light < radiation of FM radio

Given:

Wavelength, λ = 337.1 nm

Number of photons = 5.6 × 10²⁴

Energy, E = nhv = nhc/λ

Where

n = number of photons emitted

h = Planck’s constant

c = velocity of radiation

λ = wavelength of radiation

Substituting the values in the given expression of Energy [E]:

= {[5.6× 10²⁴] × [6.626× 10^-34] × [3× 10⁸ ms^-1]} / [337.1× 10^-9m]

= [1.113168] / [337.1× 10^-9m]

= 3.302× 10⁶ J

Hence, the power of the laser is 3.302× 10⁶ J.

Given:

Wavelength, λ = 616 nm = 616× 10^-9m

To find frequency:

Speed of Light = [Frequency] × [Wavelength]

We know speed of light = 3×10⁸ m/s

Frequency, v = [3×10⁸] / [616 × 10^-9]

v = 4.87×10¹⁴ Hz

To find distance travelled:

Velocity of the radiation = 3× 10⁸ m/s

Therefore, distance travelled in 30s = 30× 3× 10⁸ = 9.0× 10⁹m

To find the energy:

By Planck’s relation we have,

Energy, E = h×v

= [6.626×10^-34] × [4.87×10¹⁴]

= 3.23×10^-19 J

= 2 ev.

Therefore, the energy of the photon corresponding to light of frequency 4.87×10¹⁴ Hz is 3.23×10^-19 J.

Given:

Energy, E = 3.15 × 10^–18 J

Wavelength, λ = 600 nm

By Planck’s relation we have,

Energy, E = h×v

But we know v = [c] / [λ]

Where

c = Speed of Light

v= Frequency

λ = Wavelength

So, E = hc /λ

= [[6.626×10^-34] × [3×10⁸]] / [600×10^-9]

= [1.9878×10^-25] / [600×10^-9]

= 3.313×10^-19 J

Therefore, number of photons received = [3.15 × 10^–18] / [3.313×10^-19]

= 9.5

Therefore, the number of photons received cannot be in fraction

So, number of photons received is 9.

Given:

Period, T = 2 ns

Frequency, v = [1] / T

= [1] / [2.0 × 10^–9]

= 0.5 × 10⁹ Hz

By Planck’s relation we have,

Energy, E = n× h×v

Where

n = number of photons

h = Planck’s constant

E = [2.5× 10¹⁵] × [6.626× 10^-34] × [0.5× 10⁹]

= 8.28× 10^-10 J

Therefore, the energy of the photon is 8.28× 10^-10 J.

Given:

Wavelength, λ₁ = 589 nm

Wavelength, λ₂ = 589.6 nm

To find frequency:

Speed of Light = [Frequency] × [Wavelength]

We know speed of light = 3×10⁸ m/s

Frequency, v₁ = [3×10⁸] / [589 × 10^-9]

v₁ = 5.093×10¹⁴ Hz

Frequency, v₂ = [3×10⁸] / [589.6 × 10^-9]

V₂ = 5.088×10¹⁴ Hz

Change in Energy, ∆E = E₂-E₁ = h[v₂-v₁]

= [6.626× 10^-34] {[5.093 - 5.088] × 10¹⁴}

= [6.626× 10^-34] {[5× 10^-3] × 10¹⁴}

= 3.313× 10^-22J

Therefore, the energy difference between two excited states is 3.31× 10^-22J.

Given:

Work Function for Caesium, W = 1.9 eV

To find threshold frequency:

W = hv_o

Where

h = Planck’s constant

v_o = threshold frequency

v_o = h / w

v_o = [1.9× 1.602× 10^-19] / [6.626× 10^-34]

= [3.0438× 10^-19] / [6.626× 10^-34]

= 4.5937× 10¹⁴ Hz

Therefore, the frequency is 4.59× 10¹⁴ Hz

To find wavelength:

Speed of Light = [Frequency] × [Wavelength]

We know speed of light = 3×10⁸ m/s

Wavelength, λ = [3×10⁸] / [4.5937× 10¹⁴]

= 6.5306× 10^-7 m

Therefore, the wavelength is 6.53× 10^-7 m

Finding Kinetic Energy:

K.E of ejected electron = h[v-v_o] = hc[1/λ – 1/λ _o]

= [6.626× 3× 10^-26] [1/500× 10^-9 – 1/654× 10^-9]

= [6.626× 3× 10^-26] × {10⁹ × [154/327000]}

= [6.626× 3× 10^-26] × [468747.1289]

= 9.3177× 10^-20J

Finding Velocity of photoelectron:

We know the formula for kinetic energy which is given as follows:

1/2 mv² = kinetic Energy

Where

m = mass of electron

v = velocity of electron

v² = [2× 9.3177× 10^-20] / [9.1× 10^-31]

= [1.8654× 10^–19] / [9.1× 10^-31]

v² = 2.0478× 10¹¹

Therefore v = √ [2.0478× 10¹¹]

v = 4.525× 10⁵ m/s

Therefore, the velocity of photoelectron is 4.525× 10⁵ m/s

Let us suppose threshold wavelength to be λ nm the kinetic energy of the radiation is given as:

h (v-v_o) = 1/2 mv²

Or

hc (1/λ -1/λ_o) = 1/2 mv²

……………………………. [1]

Similarly, we can also write,

……………………………………. [2]

………………………………………. [3]

Dividing equation [3] and [1]

17.6070λ - 5λ = 8803.537 – 2000

λ = [6805.537] / [12.607]

λ = 539.8 nm

The wavelength is 540 nm.

Given:

Work Function of metal, W = 037eV

Wavelength, λ = 256.7 nm

From the Law of conservation of energy, the energy of an incident photon [E] is equal to the sum of the work function [W] of radiation and its kinetic energy [K.E] i.e.,

Energy of incident radiation [E] = hc/λ = [6.626× 10^-34] [3× 10⁸] [256.7× 10^-9]

= 7.74× 10^-19 J

Since the potential applied gives the kinetic energy to the radiation, therefore K.E of the electron = 0.35Ev

Therefore, work function = 4.83 – 0.35

= 4.48 eV

Therefore, the work function of metal is 4.48 eV

Given:

Wavelength, λ = 150 pm

Velocity, v = 1.5 × 10⁷ m s^–1

Energy, E = hv = hc/λ

Where

n = number of photons emitted

h = Planck’s constant

c = velocity of radiation

λ = wavelength of radiation

Substituting the values in the given expression of Energy [E]:

= {6.626× 10^-34] × [3× 10⁸ ms^-1]} / [150× 10^-12m]

= [1.9878× 10^-28] / [337.1× 10^-9m]

= 1.3252 × 10^-15 J

We know the formula for kinetic energy which is given as follows:

1/2 mv² = kinetic Energy

Where

m = mass of electron

v = velocity of electron

K.E. = 1/2 × {[9.1× 10^-31] × [1.5 × 10⁷]²

= 1.02375× 10^-16 J

Energy which bounded the electron to nucleus is:

= 13.25× 10^-16J-1.025× 10^-16J

=12.227× 10^-16 J

= [12.227× 10^-16]/ [1.602× 10^-19]

=7.63× 10³eV

Therefore, the energy which bounded the electron to nucleus is 7.63× 10³eV.

Given:

Wavelength, λ = 1285 nm

Frequency, v = 3.29 × 10¹⁵ [1/3² – 1/n²] (Hz)

Speed of Light = [Frequency] × [Wavelength]

We know speed of light = 3×10⁸ m/s

Frequency, v = [3×10⁸] / [1285 × 10^-9]

v = 2.3346×10¹⁴ Hz

3.29 × 10¹⁵ [1/3² – 1/n²] = 2.3346×10¹⁴

[1/3² – 1/n²] = [2.3346×10¹⁴] / [3.29 × 10¹⁵]

[1/3² – 1/n²] = 0.07427

[1/9] – 0.07427 = 1/n²

1/n² = 0.04

1/n² = 1 / 25

n² = 25

n = 5

The value of n is 5.

Given:

Radius, r₁ = 1.3225 nm

Radius, r₂ = 211.6 pm

The radius of the n^th orbit of hydrogen – like particles = 0.529n²/Z Å

Now r₁ = 1.3225 nm or 1322.5 pm = 52.9n₁² / Z

And

R₂ = 211.6pm = 52.9n₂²/Z

Taking the ratio of r1and r₂

So r₁/r₂=1322.5 / 211.6 = n₁²/n₂²

n₁²/n₂² = 6.25

n₁/n₂ = 2.5

Therefore n₂ = 2, n₁ = 5

By referring to Balmer Formula:

We know that

Wave Number

Here n₁ = 2 and n₂ = 5 and the value of R_h = 109678

Wave Number

= [109678×21] / 100

= 2303238 / 100

= 23032.38 cm^-1

We know that Wave Number = [1] / [Wavelength, λ]

Therefore, Wavelength = [1 / 23032.38]

λ = 4.3417× 10^-5 cm

Therefore, the wavelength of the light is 4.34× 10^-5 cm

The transition belongs to visible region.

Given:

Velocity of electron, v = 1.6 × 10⁶ ms^–1

Mass of electron, m = 9.1× 10^-31 kg

According to the de Broglie’s Equation:

λ = h/mv

Where,

λ = wavelength of moving particle

m = mass of electron

v= velocity of particle

h= Planck’s constant [6.62× 10^-34]

λ = {[6.62× 10^-34] / [9.1× 10^-31] × [1.6 × 10⁶]}

= {[6.62× 10^-34] / [1.456 × 10^-24]

= 4.5508× 10^-10 m

Therefore, the wavelength is 455 pm

Given:

Mass of electron, m = 1.675× 10^-27 kg

Wavelength, λ = 800 pm

According to the de Broglie’s Equation:

λ = h/mv

Where,

λ = wavelength of moving particle

m = mass of electron

v= velocity of particle

h= Planck’s constant [6.62× 10^-34]

v = {[6.62× 10^-34] / [1.675× 10^-27] × [800 × 10^-12]}

= {[6.62× 10^-34] / [1.34× 10^-36]

= 494.4776 m / s

Therefore, the velocity is 494.48 m / s.

Given:

Velocity of electron, v = 2.19 × 10⁶ ms^–1

According to the de Broglie’s Equation:

λ = h/mv

Where,

λ = wavelength of moving particle

m = mass of electron

v= velocity of particle

h= Planck’s constant [6.62× 10^-34]

λ = {[6.62× 10^-34] / [9.1× 10^-31] × [2.19 × 10⁶]}

= {[6.62× 10^-34] / [1.9929 × 10^-24]

= 3.3248× 10^-10 m

Therefore, the wavelength is 332 pm

Given:

Velocity of ball, v = 4.37 × 10⁵ ms^–1

Mass of ball, m = 0.1kg

According to the de Broglie’s Equation:

λ = h/mv

Where,

λ = wavelength of moving particle

m = mass of electron

v= velocity of particle

h= Planck’s constant [6.62× 10^-34]

λ = {[6.62× 10^-34] / [0.1] × [4.37 × 10⁵]}

= {[6.62× 10^-34] / [43.7 × 10³]

= 1.516× 10^-38 m

Therefore, the wavelength is 1.516× 10^-38 m.

By referring to Heisenberg’s uncertainty principle, we know

∆x × ∆p = h/4∏

Where,

∆x = uncertainty in position of the electron

∆p = uncertainty in momentum of the electron

∆x = 0.002nm = 2× 10^-12m [given]

Finding the value of ∆p:

∆p = [h/4π] × [1 / ∆x]

∆p = [1/0.002] × {6.626× 10^-34/4× [3.14]}

= [1 / 2× 10^-12] × {6.626×10^-34/4× 3.14}

= [5.2728×10^-35] × [5×10¹¹]

= 2.6364 × 10^-23 Jsm^-1

∆p = 2.637 × 10^-23 kgms^-1 {1 J – 1 kgms²s^-1}

Actual momentum = h / [4n× 5× 10^-11]

= [6.626× 10^-34] / [4× 3.14× 5× 10^-11]

= 1.055 × 10^-24 kg m/sec

Quantum number refers to the set of four special numbers which provide the entire information about electrons of a particular atom.

‘n’ refers to the principle quantum number

‘l’ refers to the azimuthal quantum number

‘m’ refers to the magnetic quantum number

For n = 4 and L = 2, the orbital occupied is 4d.

For n = 3 and L = 2, the orbital occupied is 3d.

For n = 4 and L = 1, the orbital occupied is 4p.

Hence, the six electrons i.e., 1, 2,3,4,5, and 6 are present in the 4d, 3d, 4p, 3d, 3p, and 4p orbital respectively.

Therefore, the increasing order of energies of the electrons is 5[3p] < 2[3d] = 4[3d] <3[4p] = 6[4p] < 1[4d].

Among the mentioned orbitals, the electrons present in 4p subshell experiences the lowest effective nuclear charge as these electrons are farthest away from the nucleus.

Nuclear charge is defined as the net positive charge experienced by an electron in the orbital of an atom exerted by the nucleus of the atom. The closer the orbital, the greater is the nuclear charge experienced by the electron [s] in it and the nuclear charge is inversely proportional to the distance of the electron from the nucleus.

(i) 3s is farther away from the nucleus as compared to 2s. Hence 2s will experience larger effective nuclear charge as compared to 3s.

(ii) In this case 4d will experience more nuclear charge as compared to 4f as 4d is more near to the nucleus.

(iii) 3f is farther away from the nucleus as compared to 3p. Hence 3p will experience larger effective nuclear charge as compared to 3f.

Nuclear charge is defined as the net positive charge experienced by an electron in the orbital of an atom exerted by the nucleus of the atom. The closer the orbital, the greater is the nuclear charge experienced by the electron [s] in it and the nuclear charge is inversely proportional to the distance of the electron from the nucleus.

Silicon has greater nuclear charge [+14] than aluminium [+13].

Hence the effective nuclear charge exerted on unpair 3p electron of silicon would be greater as compared to that of aluminium.

(a) Therefore, number of unpaired electron = 3

(b) Therefore, number of unpaired electron = 2 [as p orbital can have a maximum of 6 electrons]

(c) Therefore, number of unpaired electron = 6

(d)Therefore number of unpaired electron = 4

(e) Krypton has no unpaired electron as it is noble gas.

(a) n = 4 [Given]

For a given value of ‘n’, ‘L’ can have values from zero to [n-1].

∴ L = 0, 1, 2, 3

Thus, four sub-shells are associated with n = 4, which are s, p, d and f.

(b) Number of orbitals in the n^th shell = n²

For n = 4

Number of orbitals = 16

Each orbital has one electron with m_s = -1/2

Hence there will be 16 electrons with m_s = -1/2