Dual Nature Of Light Question 349
Question: Electrons with de-Broglie wavelength X fall on the target in an X-ray tube. The cut-off wavelength of the emitted X-rays is
Options:
A) $ {\lambda_{0}}=\frac{2mc{{\lambda }^{2}}}{h} $
B) $ {\lambda_{0}}=\frac{2h}{mc} $
C) $ {\lambda_{0}}=\frac{2m^{2}c^{2}{{\lambda }^{3}}}{h} $
D) $ {\lambda_{0}}=\lambda $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] The cut off wavelength is given by $ {\lambda_{0}}=\frac{hc}{eV} $ ?.(i) According to de Broglie equation $ \lambda =\frac{h}{p}=\frac{h}{\sqrt{2meV}} $
$ \Rightarrow {{\lambda }^{2}}=\frac{h^{2}}{2meV}\Rightarrow V=\frac{h^{2}}{2me{{\lambda }^{2}}} $ ???(ii) From (i) and (ii), $ {\lambda_{0}}=\frac{hc\times 2me{{\lambda }^{2}}}{eh^{2}}=\frac{2mc{{\lambda }^{2}}}{h} $
BETA
