Verified Solution • 2025–26 Session moderate

QNEET-2021-Ph-623: NEET 2021 Physics - Class 12 NEET-2021-Ph-623 Solution: An electromagnetic wave of wavelength '\( \)' is incident on a…

An electromagnetic wave of wavelength 'λ\lambda' is incident on a photosensitive surface of negligible work function. If 'm' mass is of photoelectron emitted from the surface has de-Broglie wavelength λd\lambda_d then: λ=(2hmc)λd2\lambda = (\frac{2h}{mc})\lambda_d^2 λ=(2mhc)λd2\lambda = (\frac{2m}{hc})\lambda_d^2 λd=(2mch)λ2\lambda_d = (\frac{2mc}{h})\lambda^2 λ=(2mch)λd2\lambda = (\frac{2mc}{h})\lambda_d^2

Multiple Choice Options

  • A λ=(2hmc)λd2\lambda = \left( {{{2h} \over {mc}}} \right){\lambda _d}^2 _ = ( m c 2 h _ ) _ d _ 2
  • B λ=(2mhc)λd2\lambda = \left( {{{2m} \over {hc}}} \right){\lambda _d}^2 _ = ( h c 2 m _ ) _ d _ 2
  • C λd=(2mch)λ2{\lambda _d} = \left( {{{2mc} \over h}} \right){\lambda ^2} _ d _ = ( h 2 m c _ ) _ 2
  • D λ=(2mch)λd2\lambda = \left( {{{2mc} \over h}} \right){\lambda _d}^2 _ = ( h 2 m c _ ) _ d _ 2

Step-by-Step Verified Solution

h c _ = k max + _ [given _ is negligible] So, h c _ = K max _ d = h 2 m K max _ K max = h 2 2 m _ d 2 ( h c _ ) = h 2 2 m _ d 2 _ _ = ( 2 m c h ) _ d 2

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