AtomsNEET Physics · Class 12 · NCERT Chapter 12

30 NEET previous-year questions on Atoms, each with the correct answer and a step-by-step solution. Filter by topic and expand any question to see how to solve it.

PYQ frequency · topic × year

17
18
19
20
21
22
23
24
Rutherford scattering
1
1
1
Bohr postulates
1
1
Bohr radius / velocity
2
2
Energy levels
1
2
1
1
1
1
Hydrogen spectrum
1
1
1
1
1
Rydberg formula
2
1
2
Excitation / ionisation
3
de Broglie standing wave
1

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All (30)
Rutherford scattering (3)
Bohr postulates (2)
Bohr radius / velocity (4)
Energy levels (7)
Hydrogen spectrum (5)
Rydberg formula (5)
Excitation / ionisation (3)
de Broglie standing wave (1)

A

-13.6/n eV

B

-13.6/n² eV

C

-13.6 n eV

D

13.6/n² eV

Solution

eV.

A

n h

B

n h/2π

C

h/n

D

2πn h

Solution

.

A

0.529 Å

B

5.29 Å

C

52.9 Å

D

0.0529 Å

Solution

a_0 = 0.529 Å.

A

UV

B

Visible

C

IR

D

Microwave

Solution

Balmer series: visible region of H spectrum.

A

13.6 eV

B

10.2 eV

C

3.4 eV

D

27.2 eV

Solution

Ionisation energy = -E_1 = 13.6 eV.

A

486 nm

B

656 nm

C

434 nm

D

410 nm

Solution

H-alpha: 656.3 nm (red).

A

-13.6 eV

B

-3.4 eV

C

-1.51 eV

D

-0.85 eV

Solution

eV.

A

Atom is mostly empty space with a dense nucleus

B

Electrons revolve in shells

C

Energy is quantised

D

Wave nature of light

Solution

Most alphas pass through, few backscatter. Atom is mostly empty; positive charge in tiny nucleus.

A

n × a_0

B

n² × a_0

C

a_0/n

D

a_0/n²

Solution

.

A

-13.6 eV

B

-27.2 eV

C

-54.4 eV

D

-108.8 eV

Solution

eV.

A

n_1 = 1

B

n_1 = 2

C

n_1 = 3

D

n_1 = 4

Solution

Lyman series: transitions ending at n_1 = 1 (UV).

A

91.2 nm

B

102.6 nm

C

121.6 nm

D

364 nm

Solution

Series limit n_2 = ∞: nm.

A

13.6 eV

B

3.4 eV

C

10.2 eV

D

17.0 eV

Solution

eV.

A

1

B

n

C

2n

D

Solution

wavelengths.

A

3

B

4

C

6

D

10

Solution

.

A

|E_n|

B

2 |E_n|

C

|E_n|/2

D

0

Solution

In Bohr orbit: KE = -E_n = +13.6/n² eV. PE = 2 E_n.

A

a_0

B

2 a_0

C

4 a_0

D

a_0/2

Solution

.

A

Lyman

B

Balmer

C

Paschen

D

Brackett

Solution

Hα (3→2), Hβ (4→2): Balmer series.

A

Charge of alpha

B

Z of target

C

KE of alpha

D

Mass of alpha

Solution

. No mass of alpha appears.

A

K = E

B

K = -E

C

K = 2E

D

K = E/2

Solution

KE = -E (PE = 2E, so total = E).

A

13.6 eV

B

27.2 eV

C

54.4 eV

D

108.8 eV

Solution

eV.

A

91.2 nm

B

121.6 nm

C

486 nm

D

656 nm

Solution

Longest = first transition: 2→1 → 121.6 nm.

A

Random

B

Quantised in angular momentum

C

Spiral

D

Elliptical

Solution

L = nℏ. Only certain orbits allowed.

A

4

B

5

C

10

D

20

Solution

.

A

-13.6 eV

B

-3.4 eV

C

-1.51 eV

D

-0.85 eV

Solution

eV.

A

Lyman

B

Balmer

C

Paschen

D

Brackett

Solution

To n_1 = 3: Paschen series (IR).

A

c/137

B

c/(137 n)

C

cn/137

D

137 c/n

Solution

for hydrogen.

A

Independent of Z

B

∝ 1/Z

C

∝ Z

D

∝ Z²

Solution

(and 1/E_k).

A

1:2

B

2:1

C

1:1

D

4:1

Solution

. v_1 : v_2 = 2 : 1.

A

Same

B

In UV

C

In visible

D

In IR

Solution

Lyman series → UV.

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