WBJEE · Physics · Dual Nature of Matter
An electron accelerated through a potential of \(10000 \mathrm{V}\) from rest has a de-Broglie wave length \(\lambda\). What should be the accelerating potential, so that the wavelength is doubled?
- A \(20000 \mathrm{V}\)
- B \(40000 \mathrm{V}\)
- C \(5000 \mathrm{V}\)
- D \(2500 \mathrm{V}\)
Answer & Solution
Correct Answer
(D) \(2500 \mathrm{V}\)
Step-by-step Solution
Detailed explanation
\(\because\) Kinetic energy of a electron due to accelerated by a potential \(V, \mathrm{KE}=e V\) \[ \frac{1}{2} m_{r} v^{2}=c V \] \(\Rightarrow \quad \frac{1 \times p^{2}}{2 m_{e}}=e V \quad[\because p=m v]\) \(\therefore\) \[ p=\sqrt{2 e V m_{e}} \] \(\because\) de-Broglie…
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