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An electron of mass $\mathbf{m}$ and a photon have same energy $\mathrm{E}$. The ratio of de-Broglie wavelengths associated with them is: A $\quad \frac{1}{c}\left(\frac{E}{2 m}\right)^{\frac{1}{2}}$ B $\quad\left(\frac{E}{2 \mathrm{~m}}\right)^{\frac{1}{2}}$ C $\quad c(2 m E)^{\frac{1}{2}}$ D $\quad \frac{1}{c}\left(\frac{2 m}{E}\right)^{1 / 2}$ (c being velocity of light)
An electron of mass $\mathbf{m}$ and a photon have same energy $\mathrm{E}$. The ratio of de-Broglie wavelengths associated with them is:
A $\quad \frac{1}{c}\left(\frac{E}{2 m}\right)^{\frac{1}{2}}$
B $\quad\left(\frac{E}{2 \mathrm{~m}}\right)^{\frac{1}{2}}$
C $\quad c(2 m E)^{\frac{1}{2}}$
D $\quad \frac{1}{c}\left(\frac{2 m}{E}\right)^{1 / 2}$
(c being velocity of light)
Physics
An electron of mass $\mathbf{m}$ and a photon have same energy $\mathrm{E}$. The ratio of de-Broglie wavelengths associated with them is:
A $\quad \frac{1}{c}\left(\frac{E}{2 m}\right)^{\frac{1}{2}}$
B $\quad\left(\frac{E}{2 \mathrm{~m}}\right)^{\frac{1}{2}}$
C $\quad c(2 m E)^{\frac{1}{2}}$
D $\quad \frac{1}{c}\left(\frac{2 m}{E}\right)^{1 / 2}$
(c being velocity of light)

Correct Option A

Solution:
As we know that,

$\lambda_{\mathrm{e}}=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{e}} \mathrm{E}}}$

For photon we evaluate the value of wavelength,

$$
\begin{aligned}
&E=\frac{h c}{\lambda_{p}} \\
&\lambda_{p}=\frac{h c}{E} \\
&\therefore \lambda_{\text {photon }}=\frac{h c}{E} \\
&\therefore \frac{\lambda_{e}}{\lambda_{p}}=\frac{h}{\sqrt{2 m E}} \times \frac{E}{h c} \\
&\frac{\lambda_{e}}{\lambda_{p}}=\frac{I}{c}\left(\frac{E}{2 m}\right)^{\frac{1}{2}}
\end{aligned}
$$

i

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