Are these Dual Nature of Matter and Radiation questions free to attempt?

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

Do I need an account to practise Dual Nature of Matter and Radiation MCQs?

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

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.

Dual Nature of Matter and Radiation NEET Questions — Free MCQ Practice

● ● ● ● ●

Which of one is correct

A.

$E^{2} = p^{2} c^{2}$

B.

$E^{2} = p^{2} c$

C.

$E^{2} = {pc}^{2}$

D.

$E^{2} = p^{2} / c^{2}$

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

Explanation

(a) Momentum $p = \frac{E}{c} \Rightarrow E^{2} = p^{2} c^{2}$

● ● ● ● ●

The work function for metals A, B and C are respectively 1.92 eV, 2.0 eV and 5 eV. According to Einstein’s equation, the metals which will emit photo electrons for a radiation of wavelength 4100 Å is/are

A.

None of these

B.

A only

C.

A and B only

D.

All the three metals

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

Explanation

(c) Energy of incident radiations (in eV)

$= \frac{12375}{4100} = 3.01 eV$
Work function of metal A and B are less than 3.01 eV , so A and B will emit photo electrons.

● ● ● ● ●

The magnitude of saturation photoelectric current depends upon

A.

Frequency

B.

Intensity

C.

Work function

D.

Stopping potential

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

Explanation

(b) The value of saturation current depends on intensity. It is independent of stopping potential

● ● ● ● ●

The light rays having photons of energy 1.8 eV are falling on a metal surface having a work function 1.2 eV. What is the stopping potential to be applied to stop the emitting electrons

A.

3 eV

B.

1.2 eV

C.

0.6 eV

D.

1.4 eV

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

Explanation

(c) Stopping potential = 1.8 eV-1.2 eV=0.6 eV

● ● ● ● ●

A photon and an electron have equal energy E. $λ_{photon} / λ_{electron}$ is proportional to

A.

$\sqrt{E}$

B.

$1 / \sqrt{E}$

C.

1/E

D.

Does not depend upon E

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

Explanation

(b) $λ_{photon} = \frac{hc}{E} and λ_{photon} = \frac{h}{\sqrt{2 mE}} \Rightarrow \frac{λ_{photon}}{λ_{electron}} = c \sqrt{\frac{2 m}{E}} \Rightarrow \frac{λ_{photon}}{λ_{electron}} \propto \frac{1}{\sqrt{E}}$

● ● ● ● ●

An image of the sun is formed by a lens of focal length of 30 cm on the metal surface of a photoelectric cell and a photoelectric current I is produced. The lens forming the image is then replaced by another of the same diameter but of focal length 15 cm. The photoelectric current in this case is

A.

$\frac{I}{2}$

B.

I

C.

2I

D.

4I

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

Explanation

● ● ● ● ●

A photon of $1.7 \times 10^{- 13}$ Joules is absorbed by a material under special circumstances. The correct statement is:

A.

Electrons of the atom of absorbed material will go the higher energy states

B.

Electron and positron pair will be created

C.

Only positron will be produced

D.

Photoelectric effect will occur and electron will be produced

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

Explanation

(b) For electron and positron pair production, minimum energy is 1.02 MeV.
Energy of photon is given $1.7 \times 10^{- 3}$ J

= $\frac{1.7 \times 10^{- 13}}{1.6 \times 10^{- 19}}$
= 1.06 MeV.
Since energy of photon is greater than 1.02 MeV,
So electron, positron pair will be created.

● ● ● ● ●

The maximum velocity of an electron emitted by light of wavelength $λ$ incident on the surface of a metal of work function $ϕ$ is

A.

${[\frac{2 (hc + λϕ)}{mλ}]}^{1 / 2}$

B.

${[\frac{2 (hc + λϕ)}{mλ}]}^{1 / 2}$

C.

${[\frac{2 (hc - λϕ)}{mλ}]}^{1 / 2}$

D.

${[\frac{2 (hλ - ϕ)}{m}]}^{1 / 2}$

Where h = Planck's constant, m = mass of electron and c = speed of light.

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

Explanation

(c) According to Einstein’s photoelectric equation

$\frac{hc}{λ} = ϕ + \frac{1}{2} {mv}^{2} \Rightarrow v = {[\frac{2 (hc - λϕ)}{mλ}]}^{1 / 2}$

● ● ● ● ●

In a photoemissive cell with executing wavelength $λ$ , the fastest electron has speed v. If the exciting wavelength is changed to 3 $λ$ /4, the speed of the fastest emitted electron will be

A.

$v {(3 / 4)}^{1 / 2}$

B.

$v {(4 / 3)}^{1 / 2}$

C.

Less than $v {(4 / 3)}^{1 / 2}$

D.

Greater than $v {(4 / 3)}^{1 / 2}$

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

Explanation

(d)

$hv - W_{0} = \frac{1}{2} {mv}_{\max}^{2} \Rightarrow \frac{hc}{λ} - \frac{hc}{λ_{0}} = \frac{1}{2} {mv}_{\max}^{2} \Rightarrow hc (\frac{λ_{0} - λ}{{λλ}_{0}}) = \frac{1}{2} {mv}_{\max}^{2} \Rightarrow v_{\max} = \sqrt{\frac{2 hc}{m} (\frac{λ_{0} - λ}{{λλ}_{0}})}$

When wavelength is $λ$ and velocity is v, then

$v = \sqrt{\frac{2 hc}{m} (\frac{λ_{0} - λ}{λ_{0} λ})}$ .....(i)

When wavelength is $\frac{3 λ}{4}$ and velocity is v', then

$v' = \sqrt{\frac{2 hc}{m} (\frac{λ_{0} - 3 λ / 4}{λ_{0} \times (3 λ / 4)})}$ .....(ii)

Divide equation (ii) by (i), we get

$\frac{v'}{v} = \sqrt{(\frac{[λ_{0} - (3 λ / 4)]}{\frac{3}{4} {λλ}_{0}}) \times \frac{{λλ}_{0}}{λ_{0} - λ}} v' = v {(\frac{4}{3})}^{1 / 2} \sqrt{\frac{[λ_{0} - (3 λ / 4)]}{λ_{0} - λ}} i . e . v' > v {(\frac{4}{3})}^{1 / 2}$

● ● ● ● ●

Photoelectric emission is observed from a metallic surface for frequencies $v_{1}$ and $v_{2}$ of the incident light rays $(v_{1} > v_{2})$ . If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio of 1:k, then the threshold frequency of the metallic surface is

A.

$\frac{v_{1} - v_{2}}{k - 1}$

B.

$\frac{{kv}_{1} - v_{2}}{k - 1}$

C.

$\frac{{kv}_{2} - v_{1}}{k - 1}$

D.

$\frac{v_{2} - v_{1}}{k}$

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

Explanation

(b) By using $hv - {hv}_{0} = K_{\max}$

$\Rightarrow h (v_{1} - v_{0}) = K_{1}$ .....(i)

And $\Rightarrow h (v_{2} - v_{0}) = K_{2}$ .....(ii)

$\Rightarrow \frac{v_{1} - v_{0}}{v_{2} - v_{0}} = \frac{K_{1}}{K_{2}} = \frac{1}{K}, Hence v_{0} = \frac{{Kv}_{1} - v_{2}}{K - 1}$

Dual Nature of Matter and Radiation MCQs for NEET — Physics Questions with Answers

Frequently Asked Questions