A) \[\frac{h}{\sqrt{2Vem}}\]
B) \[\frac{1}{c}\sqrt{\frac{2m}{Ve}}\]
C) \[\frac{1}{c}\sqrt{\frac{Ve}{2m}}\]
D) \[\frac{hc}{\sqrt{\frac{Ve}{2m}}}\]
Correct Answer: C
Solution :
Energy of accelerated electron at potential V \[{{E}_{1}}=\frac{hc}{{{\lambda }_{1}}}\] \[\Rightarrow \] \[eV=\frac{hc}{{{\lambda }_{1}}}\] Also \[{{\lambda }_{1}}=\left( \frac{hc}{eV} \right)\] ?(i) \[{{E}_{2}}=\frac{h}{mv}\] \[\therefore \] \[\frac{1}{2}m{{v}^{2}}=eV\] or \[{{v}^{2}}=\left( \frac{2eV}{m} \right)\] or \[v=\sqrt{\frac{2eV}{m}}\] \[\therefore \] \[{{\lambda }_{2}}=\frac{h}{m\sqrt{\frac{2eV}{m}}}\] or \[{{\lambda }_{2}}=\frac{h}{\sqrt{2mVe}}\] ?(ii) Dividing Eq. (ii) by Eq. (i), we have \[\frac{{{\lambda }_{2}}}{{{\lambda }_{1}}}=\frac{h/\sqrt{2meV}}{hc/eV}\] Then ratio between the de-Broglie, wavelength of electron and shortest wavelength of X-ray \[\therefore \] \[\frac{{{\lambda }_{2}}}{{{\lambda }_{1}}}=\frac{1}{c}\sqrt{\frac{eV}{2m}}\]
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