Verified Solution • 2025–26 Session moderate

QNEET-2016-Ph-1154: NEET 2016 Physics - Class 12 NEET-2016-Ph-1154 Solution: Electrons of mass m with de-Broglie wavelength \( \) fall on…

Electrons of mass m with de-Broglie wavelength λ\lambda fall on the target in an X-ray tube. The cutoff wavelength (λ0\lambda_0) of the emitted X-ray is λ0=h2mcλ22\lambda_0 = \frac{h^2mc\lambda^2}{2} λ0=2hmcλ\lambda_0 = \frac{2h}{mc\lambda} λ0=h22m2c2λ3\lambda_0 = \frac{h^2}{2m^2c^2\lambda^3} λ0=λ\lambda_0 = \lambda

Multiple Choice Options

  • A λ\lambda _ 0 = 2mcλ2h{{2mc{\lambda ^2}} \over h} h 2 m c _ 2 _
  • B λ0=2hmc{\lambda _0} = {{2h} \over {mc}} _ 0 _ = m c 2 h _
  • C λ0=2m2c2λ3h2{\lambda _0} = {{2{m^2}{c^2}{\lambda ^3}} \over {{h^2}}} _ 0 _ = h 2 2 m 2 c 2 _ 3 _
  • D λ0=λ{\lambda _0} = \lambda _ 0 _ = _

Step-by-Step Verified Solution

Kinetic energy of electrons K = p 2 2 m = ( h _ ) 2 2 m = h 2 2 m _ 2 For certain frequency, maximum wavelength that can be emitted is _ 0 which is cut off wavelength obtained at cut off frequency, E 0 = h c _ 0 Since, E = E 0 h 2 2 m _ 2 = h c _ 0 _ _ 0 = 2 m c _ 2 h

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