Chemistry XI Science
Solved Questions of Atomic Structure
Problem 1: Why Bohr’s orbits are called stationary states?
Solution: This is because the energies of orbits in which the electrons
revolve are fixed.
Problem 2: Explain why the electronic configuration of Cu is 3d104s1 and not 3d94s2.
Solution: In the 3d104s1 the d-sub shell is completely filled which is more
stable.
Problem 3: Fe3+ ion is more stable than Fe2+ ion. Why?
Solution: In Fe3+ ion 3d sub shell is half filled hence more stable
configuration.
Problem 4: Calculate the accelerating potential that must be applied to a proton beam to give it an
effective wavelength of 0.005 nm.
Solution:
v = h/mλ
ev = 1/2mv2
Putting the values we get
V = 32.85 volt
Problem 5: Give one example of isodiapheres.
Solution: Isodiapheres have same difference between the number of neutrons
and protons. For example
19K & 15P
39 31
n-p=1 m-p=1
Problem 6: Which electronic transition in Balmer series of hydrogen atom has same frequency as
that of n = 6 to n = 4 transition in He+. [Neglect reduced mass effect].
Solution :
v-He+ = RZ2 [ 1/42 - 1/62]
= 4R [ 36 - 16/36 x 16 ] = 5R/36
v-H = R x 12 [ 1/22 - 1/n2]
∴ v-He+ = v-H
5R/36 = R/4 - R/n2
On solving above equation
n2 = 9
∴n = 3
Or corresponding transition from 3 → 2 in Balmer series of hydrogen atom has same frequency as
that of 6 → 4 transition in He+.
Problem 7: Calculate ionization potential in volts of (a) He+ and (b) Li2+
Solved Questions of Atomic Structure
Problem 1: Why Bohr’s orbits are called stationary states?
Solution: This is because the energies of orbits in which the electrons
revolve are fixed.
Problem 2: Explain why the electronic configuration of Cu is 3d104s1 and not 3d94s2.
Solution: In the 3d104s1 the d-sub shell is completely filled which is more
stable.
Problem 3: Fe3+ ion is more stable than Fe2+ ion. Why?
Solution: In Fe3+ ion 3d sub shell is half filled hence more stable
configuration.
Problem 4: Calculate the accelerating potential that must be applied to a proton beam to give it an
effective wavelength of 0.005 nm.
Solution:
v = h/mλ
ev = 1/2mv2
Putting the values we get
V = 32.85 volt
Problem 5: Give one example of isodiapheres.
Solution: Isodiapheres have same difference between the number of neutrons
and protons. For example
19K & 15P
39 31
n-p=1 m-p=1
Problem 6: Which electronic transition in Balmer series of hydrogen atom has same frequency as
that of n = 6 to n = 4 transition in He+. [Neglect reduced mass effect].
Solution :
v-He+ = RZ2 [ 1/42 - 1/62]
= 4R [ 36 - 16/36 x 16 ] = 5R/36
v-H = R x 12 [ 1/22 - 1/n2]
∴ v-He+ = v-H
5R/36 = R/4 - R/n2
On solving above equation
n2 = 9
∴n = 3
Or corresponding transition from 3 → 2 in Balmer series of hydrogen atom has same frequency as
that of 6 → 4 transition in He+.
Problem 7: Calculate ionization potential in volts of (a) He+ and (b) Li2+
