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An electron accelerated through a potential of $10000 \mathrm{V}$ from rest has a de-Broglie wave length $\lambda$. What should be the accelerating potential, so that the wavelength is doubled?
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Verified Answer
The correct answer is:
$2500 \mathrm{V}$
$\because$ Kinetic energy of a electron due to accelerated by a potential $V, \mathrm{KE}=e V$
$$
\frac{1}{2} m_{r} v^{2}=c V
$$
$\Rightarrow \quad \frac{1 \times p^{2}}{2 m_{e}}=e V \quad[\because p=m v]$
$\therefore$
$$
p=\sqrt{2 e V m_{e}}
$$
$\because$ de-Broglie wavelength of a particle having momentum $p$
$$
\lambda=\frac{h}{p}
$$
According to question, $\frac{\lambda_{1}}{\lambda_{2}}=\frac{p_{2}}{p_{1}}=\frac{\sqrt{2 e V_{2} m_{c}}}{\sqrt{2 e V_{1} m_{e}}}=\sqrt{\frac{V_{2}}{V_{1}}}$
$\because$
$$
\lambda_{2}=2 \lambda_{1}
$$
$\Rightarrow \quad \frac{\lambda_{1}}{2 \lambda_{1}}=\sqrt{\frac{V_{2}}{10000}}$
$\therefore \quad$ Potential $V_{2}=\frac{10^{4}}{4}=2500 \mathrm{V}$
$$
\frac{1}{2} m_{r} v^{2}=c V
$$
$\Rightarrow \quad \frac{1 \times p^{2}}{2 m_{e}}=e V \quad[\because p=m v]$
$\therefore$
$$
p=\sqrt{2 e V m_{e}}
$$
$\because$ de-Broglie wavelength of a particle having momentum $p$
$$
\lambda=\frac{h}{p}
$$
According to question, $\frac{\lambda_{1}}{\lambda_{2}}=\frac{p_{2}}{p_{1}}=\frac{\sqrt{2 e V_{2} m_{c}}}{\sqrt{2 e V_{1} m_{e}}}=\sqrt{\frac{V_{2}}{V_{1}}}$
$\because$
$$
\lambda_{2}=2 \lambda_{1}
$$
$\Rightarrow \quad \frac{\lambda_{1}}{2 \lambda_{1}}=\sqrt{\frac{V_{2}}{10000}}$
$\therefore \quad$ Potential $V_{2}=\frac{10^{4}}{4}=2500 \mathrm{V}$
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