Blog Log in Get Started Free

Atoms MCQs for NEET — Physics Questions with Answers

Q: Are these Atoms questions free to attempt?

Yes. You can attempt every Atoms question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

Q: Do I need an account to practise Atoms MCQs?

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

Q: Are these previous year NEET questions?

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.

Practice free Atoms (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.

All Physics Chemistry Botany Zoology

Language English हिंदी

Chapter Difficulty Show only questions I answered incorrectly Clear Register free for difficulty & keyword filters

● ● ● ● ●

Frequency of the series limit of Balmer series of hydrogen atom in terms of Rydberg constant R and velocity of light C is:

$\frac{R C}{4}$

$\frac{4}{R C}$

4RC

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$\frac{1}{λ} = R (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}); C = c λ, \frac{1}{λ} = \frac{ν}{C} \frac{ν}{C} = R (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}); ν = C R (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$

For series limit n, Balmer series $n_{1} = 2, n_{2} = \infty$

$∴ ν = R C (\frac{1}{2^{2}} - \frac{1}{\infty^{2}}) = \frac{R C}{4}$

● ● ● ● ●

Orbital acceleration of electron is

$\frac{n^{2} h^{2}}{4 π^{2} m^{2} r^{3}}$

$\frac{n^{2} h^{2}}{4 n^{2} r^{3}}$

$\frac{4 n^{2} h^{2}}{π^{2} m^{2} r^{3}}$

$\frac{4 n^{2} h^{2}}{4 π^{2} m^{2} r^{3}}$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$m ν r = \frac{n h}{2 π} \Rightarrow ν = \frac{n h}{2 πmr} \Rightarrow \frac{ν^{2}}{r} = \frac{n^{2} h^{2}}{4 π^{2} m^{2} r^{3}}$

● ● ● ● ●

Consider the spectral line resulting from transition n = 2 to n = 1 in the atoms and ions given below. The shorterst wavelength is given by

Hydrogen atom

Deuterium

Singly ionised helium

Doubly ionised lithium

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$\frac{1}{λ} = \frac{Z^{2} m e^{4}}{8 ε_{0}^{2} C h^{3}} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$

Hydrogen atom Z = 1

Deutron Z = 1

Singly ionised He , Z = 2

Doubly ionised Li , Z = 3 (Gives shortest $λ$ )

● ● ● ● ●

Light is a kind of wave. The wavelength of visible light ranges from about 4000 $\overset{\circ}{A}$ to 7000 $\overset{\circ}{A}$ . Which of the following statements is false?

Ultraviolet light has a wavelength shorter than 4000 $\overset{\circ}{A}$

Infrared light has a wavelength longer than 7000 $\overset{\circ}{A}$

Red light has wavelength near about 7000 $\overset{\circ}{A}$

Violet light has wavelength near about 7000 $\overset{\circ}{A}$ .

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

Visible light has wavelength ranging from 4000 $\overset{\circ}{A}$ to 7000 $\overset{\circ}{A}$ from violet to red, $λ$ red > $λ$ violet. Ultraviolet has shorter wavelength than violet (4000 $\overset{\circ}{A}$ ). Infra red has longer wavelength than red (7000 $\overset{\circ}{A}$ ). Therefore, choice D is correct, which is a false statement.

● ● ● ● ●

An $α -$ particle of energy 5 MeV is scattered through $180 °$ by a fixed uranium nucleus. The distance of closest approach is of the order

$10^{- 10} m$

$10^{- 13} m$

$10^{- 14} m$

$10^{- 16} m$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

If $r_{0}$ is or distance of closest approach,

$K E = P E E = \frac{1}{2} m v^{2} = \frac{(Z e) (2 e)}{4 {πε}_{0} r} ∴ r = \frac{(Ze) 2 e}{4 {πε}_{0} E} = \frac{9 \times 10^{9} \times 92 \times 1.6 \times 10^{- 19} \times 2 \times 1.6 \times 10^{- 19}}{5 \times 10^{6} \times 1.6 \times 10^{- 19}} \approx 10^{- 14} m$

● ● ● ● ●

As per Bohr model, the minimum energy (in eV) required to remove the electron from the ground state of double ionized lithium (Z=3) is

1.51

13.6

40.8

122.4

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

Ionisation energy = $Z^{2} R h c (\frac{1}{{(1)}^{2}} - \frac{1}{{(\infty)}^{2}})$

$= Z^{2} R h c = {(3)}^{2} \times 13.6 e V = 122.4 e V$

● ● ● ● ●

According to Bohr's theory the moment of momentum of an electron revolving in second orbit of hydrogen atom will be [MP PET 1999; KCET 2003; VITEEE 2006]

2 $πh$

$πh$

$\frac{h}{π}$

$\frac{2 h}{π}$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

Angular momentum $L = n (\frac{h}{2 π})$

For this case n = 2, hence $L = 2 \times \frac{h}{2 π} = \frac{h}{π}$

● ● ● ● ●

The ionisation energy of hydrogen atom is 13.6 eV. Following Bohr's theory, the energy corresponging to a transition between 3rd and 4th orbit is [1992]

3.40 eV

1.51 eV

0.85

0.66 eV

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

Total energy of electron for hydrogen like atom is given by

$E_{n} = - \frac{13.6 Z^{2}}{n^{2}} ∴ E_{3} = - \frac{13.6}{3^{2}} e V [Z = 1, n = 3] = - 1.51 e V E_{4} = - \frac{13.6}{4^{2}} = - 0.85 e V ∴ E_{4} - E_{3} = 1.51 - 0.85 = 0.66 e V$

● ● ● ● ●

The ratio of the energies of the hydrogen atom in its first to second excited states is

$\frac{1}{4}$

$\frac{4}{9}$

$\frac{9}{4}$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$E_{n} = \frac{- 13.6}{n^{2}} e V F o r f i r s t e x c i t e d s t a t e, n = 2 ∴ E_{2} = \frac{- 13.6}{2^{2}} e V F o r s e c o n d e x c i t e d s t a t e, n = 3 E_{3} = \frac{- 13.6}{3^{2}} e V ∴ \frac{E_{2}}{E_{3}} = \frac{9}{4}$

● ● ● ● ●

Energy levels A, B, C of a certain atom correspond to increasing values of energy i.e. $E_{A} < E_{B} < E_{C} .$ If $λ_{1}, λ_{2}, λ_{3}$ are the wavelengths of the radiation corresponding to the transitions C to B, B to A and C to A respectively, which of the following relation is correct?

$λ_{3} = λ_{1} + λ_{2}$

$λ_{3} = \frac{λ_{1} λ_{2}}{λ_{1} + λ_{2}}$

$λ_{1} + λ_{2} + λ_{3} = 0$

$λ_{3}^{2} = λ_{1}^{2} + λ_{2}^{2}$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Ready to ace NEET?

Free access · No credit card required

Browse All Questions Create Free Account

Frequently Asked Questions

Yes. You can attempt every Atoms question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.