MHT CET · Physics · Dual Nature of Matter
When an electron is accelerated through a potential ' \(\mathrm{V}\) ', the de-Broglie wavelength associated with it is ' \(4 \lambda\) '. When the accelerating potential is increased to \(4 \mathrm{~V}\), its wavelength will be
- A \(\frac{\lambda}{4}\)
- B \(\frac{\lambda}{2}\)
- C \(\lambda\)
- D \(2 \lambda\)
Answer & Solution
Correct Answer
(B) \(\frac{\lambda}{2}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text { We know } \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{meV}}} \\ & \Rightarrow \lambda \propto \frac{1}{\sqrt{\mathrm{v}}}\end{aligned}\)
\(\begin{array}{ll} & \text { Given } \mathrm{V}=4 \mathrm{~V} \\ \therefore \quad & \lambda \propto \frac{1}{\sqrt{4 \mathrm{~V}}}=\frac{4}{2 \sqrt{\mathrm{V}}} \\ \therefore \quad & \lambda \text { reduces to } \frac{\lambda}{2}\end{array}\)
\(\begin{array}{ll} & \text { Given } \mathrm{V}=4 \mathrm{~V} \\ \therefore \quad & \lambda \propto \frac{1}{\sqrt{4 \mathrm{~V}}}=\frac{4}{2 \sqrt{\mathrm{V}}} \\ \therefore \quad & \lambda \text { reduces to } \frac{\lambda}{2}\end{array}\)
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