The de Broglie wavelength of an electron is $0.4 \times 10^{10} \mathrm{~m}$ when its kinetic energy is $1.0 \mathrm{keV}$. Its wavelength will be $1.0 \times 10^{-10} \mathrm{~m}$, when its kinetic energy is

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The de Broglie wavelength of an electron is $0.4 \times 10^{10} \mathrm{~m}$ when its kinetic energy is $1.0 \mathrm{keV}$. Its wavelength will be $1.0 \times 10^{-10} \mathrm{~m}$, when its kinetic energy is

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kritika · Answer 1 · 2021-12-14T08:45:41+0000

Ans:(D)
$$
\begin{aligned}
&\lambda=\frac{\mathrm{h}}{\mathrm{p}}, \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mk}}} \\
&\text { so } \lambda \propto \frac{1}{\sqrt{\mathrm{k}}} \\
&\Rightarrow \frac{0.4 \times 10^{-10}}{1.0 \times 10^{-10}}=\frac{\sqrt{\mathrm{k}}}{\sqrt{1}} \\
&\Rightarrow \quad 0.16 \mathrm{keV}
\end{aligned}
$$

The de Broglie wavelength of an electron is \(0.4 \times 10^{10} \mathrm{~m}\) when its kinetic energy is \(1.0 \mathrm{keV}\). Its wavelength will be \(1.0 \times 10^{-10} \mathrm{~m}\), when its kinetic energy is

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