A) \[\frac{1}{20}\]
B) \[\frac{1}{100}\]
C) 1
D) \[\frac{1}{{{10}^{4}}}\]
Correct Answer: C
Solution :
For incident electron \[{{\lambda }_{1}}=\frac{h}{p}=\frac{h}{\sqrt{2km}}=\frac{h}{\sqrt{2\,m\,Ve}}\] and shortest wavelength of \[x\]rays is \[{{\lambda }_{2}}=\frac{hc}{eV}\] \[\therefore \]\[\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\frac{1}{c}\sqrt{\frac{V}{2}.\frac{e}{m}}=\frac{1}{3\times {{10}^{8}}}\sqrt{\frac{1000\times 1.8\times {{10}^{14}}}{2}}=1\] Hence, the correction option is [c].
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