NEET · Physics · STD 12 - 11. Dual nature of radiation and matter
Electrons of mass \(m\) with de-Broglie wavelength \(\lambda\) fall on the target in an \(X-\)ray tube. The cutoff wavelength \((\lambda_0)\) of the emitted \(X-\) ray is
- A \({\lambda _0} = \frac{{2{m^2}{c^2}{\lambda ^3}}}{{{h^2}}}\)
- B \(\;{\lambda _0}\) \(= λ\)
- C \(\;{\lambda _0} = \frac{{2mc{\lambda ^2}}}{h}\)
- D \(\;{\lambda _0} = \frac{{2h}}{{mc}}\)
Answer & Solution
Correct Answer
(C) \(\;{\lambda _0} = \frac{{2mc{\lambda ^2}}}{h}\)
Step-by-step Solution
Detailed explanation
Kinetic energy of electrons \(K=\frac{p^{2}}{2 m}=\frac{(h / \lambda)^{2}}{2 m}=\frac{h^{2}}{2 m \lambda^{2}}\)
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