MHT CET · Physics · Dual Nature of Matter
The kinetic energy of an electron is increased by 2 times, then the de-Broglie wavelength associated with it changes by a factor.
- A \(\frac{1}{3}\)
- B \(\frac{1}{\sqrt{3}}\)
- C 3
- D \(\sqrt{3}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{\sqrt{3}}\)
Step-by-step Solution
Detailed explanation
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 \(\frac{1}{\sqrt{3}}\)
\(\therefore \quad \mathrm{p}=\sqrt{2 \mathrm{mE}}\)...(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 \(\frac{1}{\sqrt{3}}\)
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