# Atomic Structure

## NCERT Solution

### Part 2

Question 18: What is the energy in joules, required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit and what is the wavelength of the light emitted when the electron returns to the ground state? The ground state electron energy is -2.18 × 10-11 ergs.

Answer: 1 J = 107 ergs

So, 1 ergs = 10-7 J

Or, -2.18 × 10-11 ergs = -2.18 × 10-11 × 10-7 J

= -2.18 × 10-18 J

Energy of electron (En) = (=-2.18xx10^(-18)J)/(n^2)

So, energy in Bohr’s first orbit E1 = (=-2.18xx10^(-18)J)/1

And energy in Bohr’s 5th orbit E5 = (=-2.18xx10^(-18)J)/(25)

So, required energy ΔE = E5 - E1

=((-2.18xx10^-(18)J)/(25))-(-2.18xx10^(-18) J)

=2.18xx10^(-18)(1-1/(25))J

=2.18xx10^(-18)xx(24)/(25)J

=2.09xx10^(-18) J

Wavelength of light emitted can be calculated as follows:

Δ E = hν = (hc)/(λ)

Or, λ=(hc)/(ΔE)

=((6.626xx10^(-34)Js)xx(3xx10^8ms^(-1)))/(2.09xx10^(-18)J)

= 9.51 × 10-8 m = 951 Å

Question 19: The electron energy in hydrogen atom is given by En = (-2.18 × 10-18)/n2 J. Calculate the energy required to remove an electron completely from the n = 2 orbit. What is the longest wavelength of light in cm that can be used to cause this transition?

Answer: The required energy is the difference in energy when electron jumps from orbit n = ∞ to orbit n = 2

Energy required ΔE = E - E2

=0-(-(2.18xx10^(-18)J)/4)

= 5.45 × 10-19 J

The longest wavelength of light emitted light can be calculated as follows:

λ=(hc)/(ΔE)

=((6.626xx10^(-34)Js)xx(3xx10^8ms^(-1)))/(5.45xx10^(-19)J)

= 3.647 × 10-7 m = 3.647 × 10-5 cm

Question 20: Calculate the wavelength of an electron moving with a velocity of 2.05 × 107 m s-1.

Answer: According to de Broglie’s equation, λ=h/(mv)

Mass of electron (m) = 9.1 × 10-31 kg

Velocity of electron (v) = 2.05 × 10-7 m s-1

Planck’s constant (h) = 6.626 × 10-34 kg m2 s-1

So, λ=(6.626xx10^(-24)kg\m^2s^(-1))/((9.1xx10^(-31)kg)xx(2.05xx10^7ms^(-1)))

=(6.626xx10^(-41)xx10^(31)m)/(9.1xx2.05)

=(6.626xx10^(-10)m)/(18.655)

=3.55 × 10-11 m

Question 21: The mass of an electron is 9.1 × 10-31 kg. If its KE is 3.0 × 10-25 J, calculate its wavelength.

Answer: Velocity of electron can be calculated as follows:

KE = 1/2mv^2

Or, v^2=2K_E)m

=(2xx(3.0xx10^(-25)kg)\m^2s^(-2))/(9.1xx10^(-31)kg)

= 0.659 × 106 m2 s-2

Or, v = 8.12 × 102 m s-1

Wavelength of electron can be calculated by using de Broglie’s equation

λ=h/(mv)

=(6.626xx10^(-34)kg\m^2s^(-1))/((9.1xx10^(-31)kg)xx(8.12xx10^2ms^(-1))

=(6.626xx10^(-5)m)/(9.1xx8.12)

=(6.626xx10^(-5)m)/(73.892)

= 0.08967 × 10-7 m

= 8967 × 10-10 m = 8967 Å

Question 22: Which of the following are isoelectronic species, i.e. those having the same number of electrons?

Na+, K+, Mg2+, Ca2+, S2-, Ar

Answer: Atomic number of Na is 10 and that of Mg is 11

So, no. of electrons in Na+ = 10 – 1 = 9

No. of electrons in Mg2+ = 11 – 2 = 9

Hence, both are isoelectronic species.

Question 23:

(a) Write the electronic configuration of the following ions: (i) H- (ii) Na+ (iii) O2- (d) F-

Na+ = 1s2 2s2 2p5

O2- = 1s2 2s2 2p6

F- = 1s2 2s2 2p6

(b) What are the atomic numbers of elements whose outermost electrons are represented by (i)3s1 (ii) 2p3 and (iii) 3p5

Answer: (i) 11 (ii) 7 (iii) 17

(c) Which atoms are indicated by the following configurations? (i) [He] 2s1 (ii) [Ne] 3s2 3p3 (iii) [Ar] 4s2 3d1

Answer: (i) Li (ii) P (iii) Sc

Question 24: What is the lowest value of n that allows g orbitals to exist?

Answer: Lowest value of l where g orbital can be present = 4

So, lowest value of n for g orbital = 4 + 1 = 5

Question 25: An electron is in one of the 3d orbitals. Give the possible value of n, l and ml for this electron.

Answer: For 3d orbital n = 3, l = 2, ml = -2, -1, 0, + 1, + 2

Question 26: An atom of an element contains 29 electrons and 25 neutrons. Deduce (i) the number of protons and (ii) the electronic configuration of the element.

Answer: Number of protons = no. of electrons = 29

Electronic configuration: 1s2 2s2 2p6 3s2 3p6 3d10 4s1

Question 27: Give the number of electrons in the species H+2, H2 and O+2

Answer: H+2 = 1, H2 = 2 and O+2 = 15

Question 28:

(a) An atomic orbital has n = 3. What are the possible values of l and ml?

Answer: n = 3, l = 0, 1 and 2

For l = 0 ml = 0

For l = 1 ml = - 1, 0, + 1

For l = 2 ml = -2, -1, 0, + 1, + 2

(b) List the quantum numbers (ml and l) of electrons for 3d orbital.

Answer: For 3d orbital n = 3, l = 2, ml = -2, -1, 0, + 1, + 2

(c) Which of the following orbitals are possible? (1p, 2s, 2p and 3f)

Question 29: Using s, p, d notations, describe the orbital with the following quantum numbers. (a) n = 1, l = 0, (b) n = 3, l = 1, (c) n = 4, l = 2, (d) n = 4, l = 3.

Answer: (a) 1s, (b) 4f, (c) 3p, (d) 4d

Question 30: Explain, giving reasons, which of the following sets of quantum numbers are not possible.

(a) n = 0, l = 0, ml = 0, ms = + ½

Answer: It is not possible because value of n should be at least 1

(b) n = 1, l = 0, ml = 0, ms = - ½

(c) n = 1, l = 1, ml = 0, ms = + ½

Answer: It is not possible because l should be equal to n – 1

(d) n = 2, l = 1, ml = 0, ms = - ½

(e) n = 3, l = 3, ml = -3, ms = + ½

Answer: It is not possible because l = n -1 which is not the case here

(f) n = 3, l = 1, ml = 0, ms = + ½

Question 31: How many electrons in an atom may have the following quantum numbers?

(a) n = 4, ms = - ½

Answer: If n = 4 then no. of electrons = 2n2 = 2 × 16 = 32

Half of them will have ms = - ½

So, no. of electrons with m2 (- 1/2 ) = 16

(b) n = 3, l = 0

Answer: For l = 0 value of ml = 0 and ms = + ½ and = ½ (2 electrons)

Question 32: Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit.

mνr=(nh)/(2π)

Or, mν=(nh)/(2πr)

According to de Broglie equation

λ = h/(mν)

Or, mν = h/(λ)

Comparing these two equations, we get:

(nh)/(2πr)=h/(λ)

Or, 2π r = nλ

It is clear that circumference (2πr) of Bohr orbit for hydrogen atom is an integral multiple of de Broglie wavelength.

Question 33: What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of He+ spectrum?

1/(λ)=R_H\xx\Z^2(1/(n_1^2)-1/(n_2^2))

Or, 1/(λ)=R_H\xx4(1/4-1/(16))

For H atom

1/(λ)=R_H(1/(n_1^2)-1/(n_2^2))

Comparing these two equations, we get:

 R_H\xx4(1/4-1/(16))= R_H(1/(n_1^2)-1/(n_2^2))

Or, 1-1/4= (1/(n_1^2)-1/(n_2^2))

Or, n1 = 1 and n2 = 2

This means that transition from n = 2 to n = 1 will have same wavelength

Question 34: Calculate the energy required for the process

He+ (g) → He2+ + e-

The ionization energy for H atom in ground state is 2.18 × 10-18 J atom-1

Answer: Ionisation energy for atom is given by following equation:

En = (2.18xx10^(-18)xx\Z^2)/(n^2) J atom-1

For He+ ion, Z = 2 value of n = 1

So, En = 2.18 × 10-18 × 22 J atom-1

= 8.72 × 10-18 × 22 J atom-1