JEE Advanced · Physics · 26. Dual Nature
Electrons with de-Broglie wavelength \(\lambda\) fall on the target in an X-ray tube. The cut-off wavelength of the emitted X-rays is
- A \(\lambda_0=\frac{2 m c \lambda^2}{h}\)
- B \(\lambda_0=\frac{2 h}{m c}\)
- C \(\lambda_0=\frac{2 m^2 c^2 \lambda^3}{h^2}\)
- D \(\lambda_0=\lambda\)
Answer & Solution
Correct Answer
(A) \(\lambda_0=\frac{2 m c \lambda^2}{h}\)
Step-by-step Solution
Detailed explanation
Momentum of striking electrons, \(p=\frac{h}{\lambda}\)
\(\therefore\) Kinetic energy of striking electrons, \(K=\frac{p^2}{2 m}=\frac{h^2}{2 m \lambda^2}\)
This is also, maximum energy of \(X\)-ray photons.
Therefore, \(\quad \frac{h c}{\lambda_0}=\frac{h^2}{2 m \lambda^2}\) or \(\lambda_0=\frac{2 m \lambda^2 c}{h}\)
\(\therefore\) Correct option is (a).
\(\therefore\) Kinetic energy of striking electrons, \(K=\frac{p^2}{2 m}=\frac{h^2}{2 m \lambda^2}\)
This is also, maximum energy of \(X\)-ray photons.
Therefore, \(\quad \frac{h c}{\lambda_0}=\frac{h^2}{2 m \lambda^2}\) or \(\lambda_0=\frac{2 m \lambda^2 c}{h}\)
\(\therefore\) Correct option is (a).
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