A) 0.1
B) 0.2
C) 0.3
D) 1.0
Correct Answer: A
Solution :
For incident electron, \[\frac{1}{2}m{{v}^{2}}=eV\] or \[{{p}^{2}}=2\,meV\] \[\therefore \]de-Broglie wavelength,\[{{\lambda }_{1}}=\frac{h}{p}=\frac{h}{\sqrt{2meV}}\] Shortest X-rays wavelength,\[{{\lambda }_{2}}=\frac{hc}{eV}\] \[\therefore \] \[\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\frac{1}{c}\sqrt{\left( \frac{V}{2} \right)\left( \frac{e}{m} \right)}\] \[=\frac{\sqrt{\frac{{{10}^{4}}}{2}}\times 1.8\times {{10}^{11}}}{3\times {{10}^{8}}}=0.1\]
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