MCQ
When a metal is irradiated by light having wavelength $\lambda (\lambda < \lambda_0)$, all the photoelectron  emitted are bent in a circle of radius $r$ by a magnetic of flux density $B_0$.  Find the $\frac{1}{\lambda_0}$ where $\lambda_0$ is thershold wavelength.
  • $\frac{1}{{{\lambda _0}}} = \frac{1}{\lambda } - \frac{{{B^2}{e^2}{r^2}}}{{2mhc}}$
  • B
    $\frac{1}{{{\lambda _0}}} = \frac{1}{\lambda } + \frac{{{B^2}{e^2}{r^2}}}{{2mhc}}$
  • C
    $\frac{1}{{{\lambda _0}}} = \frac{1}{\lambda } - \frac{{{B^2}{e^2}{r^2}}}{{mhc}}$
  • D
    None of these

Answer

Correct option: A.
$\frac{1}{{{\lambda _0}}} = \frac{1}{\lambda } - \frac{{{B^2}{e^2}{r^2}}}{{2mhc}}$
a
$\mathrm{K.E.}=\frac{1}{2} \mathrm{mv}^{2}=\frac{\mathrm{hc}}{\lambda}-\frac{\mathrm{hc}}{\lambda_{0}}$

$\mathrm{r}=\frac{\mathrm{mv}}{\mathrm{eB}}=\frac{\sqrt{2 \mathrm{m}\left(\frac{\mathrm{h} c}{\lambda}-\frac{\mathrm{hc}}{\lambda_{0}}\right)}}{\mathrm{eB}}$

$\frac{1}{\lambda_{0}}=\frac{1}{\lambda}-\frac{\mathrm{e}^{2} \mathrm{B}^{2} \mathrm{r}^{2}}{2 \mathrm{mnc}}$

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