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

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Light of wavelength 500 nm is incident on a metal with work function 2.28 eV. The de-Broglie wavelength of the emitted electron is

<2.8x10^-10m

<2.8x10^-9m

>2.8x10^-9m

<2.8x10^-12m

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Explanation

(c)

As energy of photon,E=hv
E=hc/λ
$E (i n e V) = \frac{12375}{λ (i n \overset{o}{A})} = 2.48 e V$

According to Einstein's photoelectric emission,we have

KE_max=E-W=2.48-2.28=0.2eV

For de-Broglie wavelength of the emitted electron.

λ_{e min}=12.27/√KE_max(eV)

=12.27/√0.2

=27.436Å

=27.436x10^-10 m

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Light with an energy flux of 25 x 10⁴ Wm^-2 falls on a perfectly reflecting surface at normal incidence. If the surface area is 15cm² the average force exerted on the surface is

1.25x10^-6N

2.5x10^-6N

1.2x10^-6N

3.0x10^-6N

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Explanation

(b)

$E n e r g y f l u x = 25 \times 10^{4} J / s - m^{2} F o r c e o n u n i t a r e a = M o m e n t u m t r a n s f e r r e d i n u n i t t i m e o n a r e a = 2 \frac{h}{λ} = 2 \frac{E}{c} = \frac{2 \times 25 \times 10^{4} J / s - m^{2}}{3 \times 10^{8}} F o r c e o n t h e t o t a l a r e a = \frac{2 \times 25 \times 10^{4} J / s - m^{2}}{3 \times 10^{8}} \times 15 \times 10^{- 4} = 2.5 \times 10^{- 6} N$

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The wavelength λe of an electron and λp of a photon of same energy E related by:

$λ_{p} \propto {λ_{e}}^{2}$

$λ_{p} \propto λ_{e}$

$λ_{p} \propto \sqrt{λ_{e}}$

$λ_{p} \propto \frac{1}{\sqrt{λ_{e}}}$

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Explanation

(a)

$W a v e l e n g t h o f t h e e l e c t r o n : λ_{e} = \frac{h}{\sqrt{2 m E}} W a v e l e g t h o f p h o t o n : λ_{p} = \frac{h c}{E} \Rightarrow {λ_{e}}^{2} = \frac{h^{2}}{2 m E} = \frac{h^{2} λ_{p}}{2 m h c} \Rightarrow λ_{p} \propto {λ_{e}}^{2}$

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A 200W sodium street lamp emits yellow light of wavelength $0.6 μ m .$ Assuming it to be 25% efficient in converting electrical energy to light, the number of photons of yellow light it emits per second is

$1.5 \times 10^{20}$

$6 \times 10^{18}$

$62 \times 10^{20}$

$3 \times 10^{19}$

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Explanation

Efficient power P = $\frac{N}{t} \times \frac{h c}{λ} = 200 \times 0.25$

$\frac{N}{t} = 50 \times \frac{λ}{h c} = 1.5 \times 10^{20}$

$= \frac{50 \times 0.6 \times 10^{- 6}}{6.6 \times 10^{- 34} \times 3 \times 10^{8}}$

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An $α ‐$ particle moves in a circular path of radius 0.83 cm in the presence of a magnetic field of 0.25 $W b / m^{2} .$ The de-Broglie wavelength associated with the particle will be

$1 \overset{0}{A}$

$0.1 \overset{0}{A}$

$10 \overset{0}{A}$

$0.01 \overset{0}{A}$

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Explanation

We knows

$R = \frac{m v}{q B}$

and $λ = \frac{h}{m v}$

$λ = \frac{h}{q B R}$

$= \frac{6.6 \times 10^{- 34}}{2 \times 1.6 \times 10^{- 19} \times 0.83 \times 10^{- 2} \times 0.25}$

$= 0.01 \overset{0}{A}$

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If the momentum of an electron is changed

by p, then the de-Broglie wavelength

associated with it changes by 0.5%. The

initial momentum of the electron will be

200p

400p

$\frac{P}{200}$

100p

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Explanation

de-Broglie wavelength

$λ = \frac{h}{p}$

Here $|\frac{∆ λ}{λ}| = |\frac{∆ p}{p}|$

$\frac{0.5}{100} = \frac{P}{p_{i}} P_{i} = \frac{100}{0.5} p P_{i} = 200 p$

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In the Davisson and Germer experiment, the

velocity of electrons emitted from the electron

gun can be increased by

increasing the filament current

decreasing the filament current

decreasing the potential difference between

the anode and filament

increasing the potential difference between

the anode and filament

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Explanation

In the Davisson and Germer experiment, the

velocity of the electron emitted from the electron

gun can be increased by increasing the potential

difference between the anode and filament.

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A radioactive nucleus of mass M emits a photon

of frequency $ν$ and the nucleus recoils. The recoil

energy will be:

$h^{2} ν^{2} / 2 {Mc}^{2}$

zero

h $ν$

${Mc}^{2} - hν$

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Explanation

Momentum of photon

$p = \frac{hν}{c}$

Hence, Recoil energy

$E = \frac{P^{2}}{2 M} E = \frac{{(\frac{hν}{c})}^{2}}{2 M}$

or $h = \frac{h^{2} ν^{2}}{2 M c^{2}}$

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In photoelectric emission process from a metal

of work function 1.8 eV, the kinetic energy of most

energetic electrons is 0.5 eV. The corresponding

stopping potential is

1.2 V

0.5 V

2.3 V

1.8 V

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Explanation

Stopping potential= Maximum KE

eV= ${KE}_{\max}$

so, option (b) is correct.

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Light of two different frequencies whose

photons have energies 1 eV and 2.5 eV

respectively illuminate a metallic surface

whose function is 0.5 eV successively.

Ratio of maximum speeds of emitted

electrons will be

1:2

1:1

1:5

1:4

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Explanation

Kinetic energy

KE= $ϕ - ϕ_{0}$

Here, ${KE}_{1} = 1 - 0.5 = 0.5 eV {KE}_{2} = 2.5 - 0.5 = 2 eV$

so, $\frac{{KE}_{1}}{{KE}_{2}} = \frac{0.5}{2} = \frac{1}{4}$

or $\frac{v_{1}^{2}}{v_{2}^{2}} = \frac{1}{4}$

or $\frac{v_{1}}{v_{2}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$

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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.

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