MCQ
light of wavelength $\lambda$ strikes a photo-sensitive surface and electrons are ejected with kinetic energy $E$. If the kinetic energy is to be increased to $2\ E$, the wavelength must be changed to $\lambda^{\prime}$ where
  • A
    $\lambda^{\prime}=\frac{\lambda}{2}$
  • B
    $ \lambda^{\prime}=2 \lambda$
  • $\frac{\lambda}{2}<\lambda^{\prime}<\lambda$
  • D
    $\lambda^{\prime}>\lambda$

Answer

Correct option: C.
$\frac{\lambda}{2}<\lambda^{\prime}<\lambda$
$ E=\frac{h c}{\lambda}-W_0 \text { and } 2 E=\frac{h c}{\lambda^{\prime}}-W_0$
$ \Rightarrow \frac{\lambda^{\prime}}{\lambda}=\frac{E+W_0}{2 E+W_0} \Rightarrow \lambda^{\prime}=\lambda (\frac{1+W_0 / E}{2+W_0 / E})$
Since $\frac{(1+W_0 / E)}{(2+W_0 / E)}>\frac{1}{2}$ so $\lambda^{\prime}>\frac{\lambda}{2}$

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