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In photoelectric effect, initially when energy of electrons emitted is $E_{o}$, de-Broglie wavelength associated with them is $\lambda_{0}$. Now, energy is doubled then associated de-Broglie wavelength $\lambda^{\prime}$ is
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The correct answer is:
$\lambda^{\prime}=\frac{\lambda_{o}}{\sqrt{2}}$
de-Broglie wavelength is given by
$\begin{array}{l}
\lambda=\frac{\mathrm{h}}{\mathrm{p}}, \text { where } \mathrm{h}=\text { Planck's constant and } \\
\mathrm{p}=\text { momentum }
\end{array}$
Also, energy (E) and momentum are related as
$\begin{array}{l}
\mathrm{E}=\frac{\mathrm{p}^{2}}{2 \mathrm{~m}} \\
\therefore \quad \mathrm{p}=\sqrt{2 \mathrm{mE}}
\end{array}$
$\therefore \quad \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}} \times \frac{1}{\sqrt{\mathrm{E}}}$ as $\mathrm{h}$ and $\mathrm{m}$ are constants
Hence, $\frac{\lambda_{0}}{\lambda^{\prime}}=\sqrt{\frac{\mathrm{E}^{\prime}}{\mathrm{E}}}=\sqrt{\frac{2 \mathrm{E}}{\mathrm{E}}}=\sqrt{2}$
$\therefore \quad \lambda^{\prime}=\frac{\lambda_{0}}{\sqrt{2}}$
$\begin{array}{l}
\lambda=\frac{\mathrm{h}}{\mathrm{p}}, \text { where } \mathrm{h}=\text { Planck's constant and } \\
\mathrm{p}=\text { momentum }
\end{array}$
Also, energy (E) and momentum are related as
$\begin{array}{l}
\mathrm{E}=\frac{\mathrm{p}^{2}}{2 \mathrm{~m}} \\
\therefore \quad \mathrm{p}=\sqrt{2 \mathrm{mE}}
\end{array}$
$\therefore \quad \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}} \times \frac{1}{\sqrt{\mathrm{E}}}$ as $\mathrm{h}$ and $\mathrm{m}$ are constants
Hence, $\frac{\lambda_{0}}{\lambda^{\prime}}=\sqrt{\frac{\mathrm{E}^{\prime}}{\mathrm{E}}}=\sqrt{\frac{2 \mathrm{E}}{\mathrm{E}}}=\sqrt{2}$
$\therefore \quad \lambda^{\prime}=\frac{\lambda_{0}}{\sqrt{2}}$
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