MHT CET · Physics · Dual Nature of Matter
Electrons of mass \(m\) with de-Broglie wavelength \(\lambda\) fall on the target. The cut-off wavelength \(\lambda_0\) is equal to [ \(h=\) Planck's constant, \(C=\) velocity of light \(]\)
- A \(\frac{2 m c \lambda^2}{h}\)
- B \(\frac{m c \lambda}{h}\)
- C \(\frac{2 h}{m c \lambda^2}\)
- D \(\frac{2 m c \lambda}{h}\)
Answer & Solution
Correct Answer
(A) \(\frac{2 m c \lambda^2}{h}\)
Step-by-step Solution
Detailed explanation
Using de-Broglie equation \(\lambda=\frac{h}{p}\) where \(p=\sqrt{2 m E}\)
\(\Rightarrow \lambda=\frac{h}{\sqrt{2 m E}}\)
Energy of the X-ray emitted \(E=\frac{h c}{\lambda_0}\)
\(\therefore \lambda=\frac{h}{\sqrt{2 m \times \frac{h c}{\lambda_0}}} \Rightarrow \lambda_0=\frac{2 m c \lambda^2}{h}\)
\(\Rightarrow \lambda=\frac{h}{\sqrt{2 m E}}\)
Energy of the X-ray emitted \(E=\frac{h c}{\lambda_0}\)
\(\therefore \lambda=\frac{h}{\sqrt{2 m \times \frac{h c}{\lambda_0}}} \Rightarrow \lambda_0=\frac{2 m c \lambda^2}{h}\)
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