The kinetic energy of an electron is increased by 2 times, then the de-Broglie wavelength associated with it…
The kinetic energy of an electron is increased by 2 times, then the de-Broglie wavelength associated with it changes by a factor.
- $\frac{1}{3}$
- $\frac{1}{\sqrt{3}}$
- 3
- $\sqrt{3}$
Solution
Kinetic Energy, $E=\frac{1}{2} \mathrm{mv}^2=\frac{1}{2} \frac{(\mathrm{mv})^2}{\mathrm{~m}}=\frac{\mathrm{p}^2}{2 \mathrm{~m}}$
$\therefore \quad \mathrm{p}=\sqrt{2 \mathrm{mE}}$...(i)
De-Broglie wavelength,
$\begin{aligned}
& \lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}} \\
& \Rightarrow \lambda \propto \frac{1}{\sqrt{\mathrm{E}}}
\end{aligned}$
...[From(i)]
$\therefore \quad$ When kinetic energy is increased by 2 times, $\mathrm{E}_2=\mathrm{E}_1+2 \mathrm{E}_1=3 \mathrm{E}_1$
$\therefore \quad \lambda$ changes by factor of $\frac{1}{\sqrt{3}}$
Asked in: MHT CET 2024 (04 May Shift 2)
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