A) 1
B) 0.1
C) 0.2
D) 0.3
Correct Answer: B
Solution :
For the incident electron \[\frac{1}{2}m{{v}^{2}}=eV\]or\[{{p}^{2}}=2meV\] \[\therefore \]de-Broglie wavelength\[{{\lambda }_{1}}=\frac{h}{p}=\frac{h}{\sqrt{2meV}}\] Shortest \[X-\]ray 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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