Blog Log in Get Started Free

Physics MCQs for NEET — Practice Questions with Answers

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

All Physics Chemistry Botany Zoology

Chapter Language Difficulty Show only questions I answered incorrectly Register free to filter questions

● ● ● ● ●

If the momentum of a photon is p, then its frequency is

$\frac{ph}{c}$

$\frac{pc}{h}$

$\frac{mh}{c}$

$\frac{mc}{h}$

where m is the rest mass of the photon

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

Explanation

(b) $p = \frac{E}{c} = \frac{hv}{c} \Rightarrow v = \frac{pc}{h}$

● ● ● ● ●

An AIR station is broadcasting the waves of wavelength 300 metres. If the radiating power of the transmitter is 10 kW, then the number of photons radiated per second is

$1.5 \times 10^{29}$

$1.5 \times 10^{31}$

$1.5 \times 10^{33}$

$1.5 \times 10^{35}$

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

Explanation

(b) $P = \frac{W}{t} = \frac{nhc}{λt} \Rightarrow (\frac{n}{t}) = \frac{Pλ}{hc} = \frac{10 \times 10^{3} \times 300}{6.6 \times 10^{- 34} \times 3 \times 10^{8}} = 1.5 \times 10^{31}$

● ● ● ● ●

The energy of a photon is E = hv and the momentum of photon $p = \frac{h}{λ}$ , then the velocity of photon will be

E/p

${(\frac{E}{P})}^{2}$

$3 \times 10^{8} m / s$

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

Explanation

(a) Momentum of photon $p = \frac{E}{c}$
$\Rightarrow$ Velocity of photon $c = \frac{E}{P}$

● ● ● ● ●

The approximate wavelength of a photon of energy 2.48 eV is

500 Å

5000 Å

2000 Å

1000 Å

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

Explanation

(b) By using

$E (eV) = \frac{12375}{λÅ} \Rightarrow λ = \frac{12375}{2.48} = 4989.9 Å \approx 5000 Å$

● ● ● ● ●

Energy of a quanta of frequency $10^{15}$ Hz and $h = 6.6 \times 10^{- 34} J - \sec$ will be

$6.6 \times 10^{- 19} J$

$6.6 \times 10^{- 12} J$

$6.6 \times 10^{- 49} J$

$6.6 \times 10^{- 41} J$

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

Explanation

(a) $E = hv = 6.6 \times 10^{- 34} \times 10^{15} = 6.6 \times 10^{- 19} J$

● ● ● ● ●

Momentum of a photon of wavelength $λ$ is

$\frac{h}{λ}$

Zero

$\frac{hλ}{c^{2}}$

$\frac{hλ}{c}$

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

Explanation

(a) Since $hv = {mc}^{2}$ , hence $p = mc = \frac{hv}{c} = \frac{h}{λ}$

● ● ● ● ●

Wavelength of a 1 keV photon is $1.24 \times 10^{- 9} m$ . What is the frequency of 1 MeV photon

$1.24 \times 10^{15} Hz$

$2.4 \times 10^{20} Hz$

$1.24 \times 10^{18} Hz$

$2.4 \times 10^{23} Hz$

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

Explanation

(b) $E = hv \Rightarrow v = \frac{E}{h} = \frac{1 \times 10^{6} \times 1.6 \times 10^{- 19}}{6.6 \times 10^{- 34}} = 2.4 \times 10^{20} Hz$

● ● ● ● ●

What is the momentum of a photon having frequency $1.5 \times 10^{13} Hz$ ?

$3.3 \times 10^{- 29} kg m / s$

$3.3 \times 10^{- 34} kg m / s$

$6.6 \times 10^{- 34} kg m / s$

$6.6 \times 10^{- 30} kg m / s$

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

Explanation

(a) $p = \frac{hv}{c} = \frac{6.6 \times 10^{- 34} \times 1.5 \times 10^{13}}{3 \times 10^{8}} = 3.3 \times 10^{- 29} kg - m / \sec$

● ● ● ● ●

The energy of a photon of light of wavelength 450 nm is

$4.4 \times 10^{- 19} J$

$2.5 \times 10^{- 19} J$

$1.25 \times 10^{- 17} J$

$2.5 \times 10^{- 17} J$

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

Explanation

(a) $E = \frac{hc}{λ} = \frac{6.62 \times 10^{- 34} \times 3 \times 10^{8}}{450 \times 10^{- 9}} = 4.4 \times 10^{- 19} J$

● ● ● ● ●

Frequency of photon having energy 66 eV is

$8 \times 10^{- 15} Hz$

$12 \times 10^{- 15} Hz$

$16 \times 10^{15} Hz$

None of these

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

Explanation

(c) $E = hv \Rightarrow v = \frac{E}{h} = \frac{66 \times 1.6 \times 10^{- 19}}{6.6 \times 10^{- 34}} = 16 \times 10^{15} Hz$

Ready to ace NEET?

Free access · No credit card required

Browse All Questions Create Free Account

Frequently Asked Questions

Yes. You can attempt every Physics 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.