JEE Main Previous Year Solved Questions on Atomic Structure

1. If the kinetic energy of an electron is increased four times, the wavelength of the de-Broglie wave associated with it would become

(1) Two times
(2) Half
(3) One fourth
(4) Four times


Solution:

The wavelength λ is inversely proportional to the square root of kinetic energy. So if KE is increased 4 times, the wavelength becomes half.
λ∝1/√KE
Hence option (2) is the answer.

2. Calculate the wavelength (in nanometer) associated with a proton moving at 1.0×103ms-1 (Mass of proton = 1.67×10-27kg and h = 6.63×10-34Js)

(1) 2.5 nm
(2) 14.0 nm
(3) 0.032 nm
(4) 0.40 nm


Solution:

Given mp = 1.67×10-27kg
h = 6.63×10-34Js
v = 1.0×103ms-1
We know wavelength λ = h/mv
∴λ = 6.63×10-34/(1.67×10-27 × 1.0×103)
= 3.97×10-10
≈ 0.04nm
Hence option (4) is the answer.

3. The radius of the second Bohr orbit for hydrogen atom is :
(Planck’s constant, h = 6.262×10-34Js: Mass of electron = 9.1091×10-31kg; Charge of electron e = 1.60210×10-19C; permittivity of vacuum ε0 = 8.854185×10-12kg-1m-3A2)


(1) 1.65 A
(2) 4.76 A
(3) 0.529 A
(4) 2.12 A



Solution:

Radius of nth Bohr orbit in H atom = 0.53 n2/Z
For hydrogen Z = 1
Radius of 2nd Bohr orbit in H atom = 0.53 ×22/1 = 2.12
Hence option (4) is the answer.

4. The energy required to break one mole of Cl–Cl bonds in Cl2 is 242 kJ mol-1. The longest wavelength of light capable of breaking a single Cl–Cl bond is (C = 3×108 m/s and NA = 6.02×1023 mol-1)


(1) 494 nm
(2) 594 nm
(3) 640 nm
(4) 700 nm




Solution:

We have B.E = 242KJ/Mol
E = hcNA/λ
∴ λ = hcNA/E
= 3×108×6.626×10-34×6.02×1023/(242×103)
= 0.494×10-3×103
= 494 nm
Hence option (1) is the answer.

5. Ionisation energy of He+ is 19.6×10-18J atom-1. The energy of the first stationary state (n = 1) of Li2+ is


(1)8.82×10-17 J atom-1
(2) 4.41×10-16 J atom-1
(3) -4.41×10-17 J atom-1
(4) -2.2×10-15 J atom-1



Solution:

Given I.E = 19.6×10-18
I.E ∝ z2
(I.E) Li2+/He+ = (9/4)×19.6×10-18
= -4.41×10-17
Hence the option (3) is the answer.

6. The frequency of light emitted for the transition n = 4 to n = 2 of He+ is equal to the transition in H atom corresponding to which of the following



(1) n = 3 to n = 1
(2) n = 2 to n = 1
(3) n = 3 to n = 2
(4) n = 4 to n = 3




Solution:

E = 13.6×4[(¼)-(1/16)]
= 10.2
E = hν
ν = 10.2/h
E = 13.6(1)[(1/n12-1/n22)]
10.2 = 13.6[(1/n12-1/n22)]
102/136 = (n22-n12)/n12n22
Substitute the given options and find n1 and n2
51/68 = (n22-n12)/n12n22
0.75 = (4-1)4 = ¾ = 0.75
Hence option (2) is the answer.

7. Based on the equation ΔE = -2.0×10-18 J (1/n2 2– 1/n12) the wavelength of the light that must be absorbed to excite hydrogen electron from level n = 1 to level n= 2 will be (h = 6.625×10-34 Js, C = 3×108 ms-1)


(1) 2.650×10-7m
(2) 1.325×10-7m
(3) 1.325×10-10m
(4) 5.300×10-10m




Solution:

ΔE = -2.0×10-18 J (1/n2 2– 1/n12)
= -2.0×10-18(1/22 – 1/12)
= -2.0×10-18(1/4 – 1/1)
= -2.0×10-18(-3/4)
= 1.5×10-18
Also ΔE = hc/λ
So λ = hc/ΔE
= 6.625×10-34 × 3×108 /1.5×10-18
=13.25×10-8
= 1.325×10-7m
Hence option (2) is the answer.

8. The de Broglie wavelength of a car of mass 1000 kg and velocity 36 km/hr is :
(h = 6.63×10-34 Js)



(1) 6.626×10-31 m
(2) 6.626×10-34 m
(3) 6.626×10-38 m
(4) 6.626×10-30 m




Solution:

Given h = 6.63×10-34 J/s
m = 1000 kg
v = 36 km/hr = 36×103/(60×60) m/s = 10m/s
λ = h/mv λ = h/mv
= 6.63×10-34 /1000×10
= 6.63×10-38 m
Hence option (3) is the answer.

9. If the binding energy of the electron in a hydrogen atom is 13.6 eV, the energy required to remove the electron from the first excited state of Li++ is



(1) 13.6 eV
(2) 30.6 eV
(3) 122.4 eV
(4) 3.4 eV




Solution:

B.E = 13.6×Z2/n2, Z is the atomic number and n is the orbital quantum number. For Li++ , Z = 3 and n = 2 for the first excited state.
B.E = 13.6×32/22
= 30.6 ev
Hence option (2) is the answer.

10. According to Bohr’s theory, the angular momentum of an electron in 5th order orbit is



(1) 25 h/π
(2) 1.0 h/π
(3) 10 h/π
(4) 2.5 h/π




Solution:

n = 5 So angular momentum, = nh/2π
= 5h/2π
= 2.5 h/π
Hence option (4) is the answer.

11. The de Broglie wavelength of a tennis ball of mass 60g moving with a velocity of 10m/s is approximately ( Planck’s constant, h = 6.63×10-34 Js)
(1) 10-31 m
(2) 10-16 m
(3) 10-25 m
(4) 10-33 m

Solution:

Given m = 60 g
v = 10 m/s
λ = h/mv
= 6.6×10-34/(60×10-3×10) = 10-33 m
Hence option (4) is the answer.

12. In a hydrogen atom, if energy of an electron in the ground state is 13.6 eV, then that in the 2nd excited state is
(1) 1.51 eV
(2) 3.4 eV
(3) 6.04 eV
(4)13.6 eV

Solution:

The 3rd energy level is the 2nd excited state.
n=3
En = 13.6/n2 = 13.6/9 = 1.5 eV
Hence option (1) is the answer.

13. In Bohr series of lines of hydrogen spectrum, the third line from the red end corresponds to which one of the following inter-orbit jumps of the electron for Bohr orbits in an atom of hydrogen
(1) 5 → 2
(2) 4 → 1
(3) 2→ 5
(4) 3→ 2

Solution:

The lines falling in the visible spectrum includes Balmer series. So the third line would be n1 = 2 and n2 = 5. Thus the transition is 5 → 2
Hence option (1) is the answer.

14. Which of the following sets of quantum numbers is correct for an electron present in 4f orbital?
(1) n = 4, l = 3, m = +4, s = +½
(2) n = 3, l = 2, m = -2, s = +½
(3) n = 4, l = 3, m = +1, s = +½
(4) n = 4, l = 4, m = -4, s = -½

Solution:

For 4f orbital, n = 4 and l = 3.
Values of m = -3, -2, -1, 0, +1, +2, +3
Hence option (3) is the answer.

15. The number of d-electrons retained in Fe2+ (At.no. of Fe = 26) ion is
(1) 4
(2) 5
(3) 6
(4) 3

Solution:

Configuration of Fe2+ = 3d6 4s0
Hence option (3) is the answer.

16. Which of the following statements in relation to the hydrogen atom is correct ?
(1)3s orbital is lower in energy than 3p orbital
(2)3p orbital is lower in energy than 3d orbital
(3)3s and 3p orbitals are of lower energy than 3d orbital
(4)3s, 3p and 3d orbitals all have the same energy

Solution:

Auf-bau principle is not applicable for Hydrogen atom.
Hence option (4) is the answer.

17. Which of the following sets of quantum numbers represents the highest energy of an atom?
(1)n=3, l =2, m=l, s= +½
(2)n=3, l =2, m=l, s= +½
(3)n=4, l =0, m=0, s= +½
(4)n=3, l =0, m=0, s= +½

Solution:

Maximum value of (n +l) represents the high­est energy of the orbital.
Hence option (2) is the answer.

18. The outer electron configuration of Gd (Atomic no. 64) is
(1) 4f4 5d4 6s2
(2) 4f7 5d1 6s2
(3) 4f3 5d5 6s2
(4) 4f8 5d0 6s2

Solution:

Gd shows half filled f7 configuration.
Hence option (2) is the answer.