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Dual Nature of Radiation and Matter question

2021 · Q149
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Dual Nature of Radiation and Matter question

2021 · Q149

NEETPhysicsDual Nature of Radiation and MatterMCQ+4 / −1
An electromagnetic wave of wavelength 'λ\lambdaλ' is incident on a photosensitive surface of negligible work function. If 'm' mass is of photoelectron emitted from the surface has de-Broglie wavelength λ\lambdaλd, then :
  1. A
    λ=(2hmc)λd2\lambda = \left( {{{2h} \over {mc}}} \right){\lambda _d}^2λ=(mc2h​)λd​2
  2. B
    λ=(2mhc)λd2\lambda = \left( {{{2m} \over {hc}}} \right){\lambda _d}^2λ=(hc2m​)λd​2
  3. C
    λd=(2mch)λ2{\lambda _d} = \left( {{{2mc} \over h}} \right){\lambda ^2}λd​=(h2mc​)λ2
  4. D
    λ=(2mch)λd2\lambda = \left( {{{2mc} \over h}} \right){\lambda _d}^2λ=(h2mc​)λd​2
View written solutionFree

Correct answer: D

hcλ=kmax⁡+ϕ{{hc} \over \lambda } = {k_{\max }} + \phi λhc​=kmax​+ϕ [given ϕ\phiϕ is negligible]

So, hcλ=Kmax⁡{{hc} \over \lambda } = {K_{\max }}λhc​=Kmax​

λd=h2mKmax⁡⇒Kmax⁡=h22mλd2{\lambda _d} = {h \over {\sqrt {2m{K_{\max }}} }} \Rightarrow {K_{\max }} = {{{h^2}} \over {2m\lambda _d^2}}λd​=2mKmax​​h​⇒Kmax​=2mλd2​h2​

(hcλ)=h22mλd2⇒λ=(2mch)λd2\left( {{{hc} \over \lambda }} \right) = {{{h^2}} \over {2m\lambda _d^2}} \Rightarrow \lambda = \left( {{{2mc} \over h}} \right)\lambda _d^2(λhc​)=2mλd2​h2​⇒λ=(h2mc​)λd2​

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