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

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de - Broglic wavelength of electron in nth Bohr orbit is............

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Explanation

According to de Broglie hypothesis, the wavelength of an electron in the nth Bohr orbit is given by: $$ ext{wavelength} = rac{2 ext{π}r}{n} $$ Here, $r$ is the radius of the nth orbit and $n$ is the principal quantum number. This formula is derived from the quantization condition of angular momentum in Bohr's model of the atom, where $mvr = n rac{h}{2 ext{π}}$. Therefore, the correct option is: $$ rac{2 ext{π}r}{n} A^ ext{∘}$$

In photo electric effect, if threshold wave length of a metal is $ 5000 A^ \circ $ work function of this metal is....................eV. $ ( h = 6.6 \times 10^{-34} J.s , c = 3 \times 10^8 m/s , 1 eV - 1.6 \times 10^{-19} J.s ) $

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Explanation

$ \phi = { hc \over \lambda_0 e } = { 6.62 \times 10^{-34 } \times 3 \times 10^8 \over 5 \times 10^{-7} \times 1.6 \times 10^{-19} }$ = 2.48 eV

Photo senstive surface is incident by light having frequecy 3 times its threshold frequency. In this condition, if frequency of light is made half and intensity of light is made double, magnitude of photo electric current becomes

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If ratio of threshold frequencies of two metals is 1 : 3, ratio of their work functions is.............

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Explanation

$ { \phi_1 \over \phi_2 } = { hf_01 \over hf_02} = { f_01 \over f_03} $ $ \therefore { \phi_1 \over \phi_2 } = {1 \over 3 } $

It work function of Na and Fe are 2.5 eV and 5eV respectively ratio of their threshold frequencies.........................

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Explanation

$ \phi = hf_0 $ $ \therefore \phi \,\alpha f_0 $ $ \therefore {(f_0)_{Na} \over (f_0)_{Fe}} = { \phi _{Na} \over \phi_{Fe} } = { 2.5 \over 5 } = {1 \over 2 } $ = 1:2

If electron is accelerated under the effect of 200V p.d., its kinetic energy = ...................

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Explanation

$ { 1 \over 2 } mv^2 max = eV = 1.6 \times 10^{-19} \times 200 = 3.2 \times 10^{-17} J $

In quantum mechaincs, a particle

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Explanation

In quantum mechanics, a particle can be described as a group of harmonic waves. This concept is fundamental to the wave-particle duality of matter, where particles exhibit both wave-like and particle-like properties. The group of waves forms a wave packet that represents the particle, with the superposition of waves leading to a localized wave packet that corresponds to the particle's position.

The de-Broglie wavelength of a proton and $ \alpha $ - particle is same. The ratio of their velocities will be..............( $ \alpha $ particle is the He-nucleus, having two protonsandtwo neutrons. Thus, its mass $M_\infty = 4 m_p $ where $m_p$ is the mass of the proton.)

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Explanation

$ \lambda_p = \lambda_\alpha $ $ \therefore { h \over m_p v_\alpha } = { h \over m_\alpha v_\alpha } $ $ \therefore { v_p \over v_\alpha} = { m_\alpha \over m_p} $ $ m_\alpha = 4m_p $ $ { v_p \over v_\alpha } = { 4m_p \over m_p} = 4 $ $ \therefore {v_p \over v_\alpha} = 4:1 $

The de-Broglie wavelength associated with a particle with rest mass $m_0$ and moving with speed of light in vacuum is..................

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Explanation

$ \lambda = { h \over mv } $ $ m = { m_0 \over \sqrt {1 - { v^2 \over c^2}} }$ $ \lambda = { h \times \sqrt { 1 - { v^2 \over c^2 } \over m_0 v }} $ $ v =c , \lambda = { h \times \sqrt { 1 - { c^2 \over c^2 } \over m_0 v }} $ $ \therefore \lambda = 0 $

A proton and electron are lying in a box having unpenetrable walls, the ratio of uncertainty in their velocities are........( $m_e $= mass of electron and $m_p $= mass of proton.)

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