A) \[{{\lambda }_{o}}=\frac{2mc{{\lambda }^{2}}}{h}\]
B) \[{{\lambda }_{o}}=\frac{2h}{mc}\]
C) \[{{\lambda }_{o}}=\frac{2{{m}^{2}}{{c}^{2}}{{\lambda }^{3}}}{{{h}^{2}}}\]
D) \[{{\lambda }_{o}}=\lambda \]
Correct Answer: A
Solution :
[a] Let "K" be the kinetic energy of the incident electron its linear momentum \[p=\sqrt{2mk}\] De-broglie wavelength \[\lambda =\frac{h}{p}=\frac{h}{\sqrt{2mk}}\,\,\,or\,\,k=\frac{{{h}^{2}}}{2m{{\lambda }^{2}}}\,\,\,......(1)\] Cut off wavelength of the emmited x-ray \[k=\frac{hc}{{{\lambda }_{o}}}\,\,\,\,.....(2)\left[ value\,\,of\,\,k=\frac{{{h}^{2}}}{2m{{\lambda }^{2}}} \right]\] \[\frac{{{h}^{2}}}{2m{{\lambda }^{2}}}=\frac{hc}{{{\lambda }_{o}}}\] \[\Rightarrow {{\lambda }_{o}}=\frac{2m{{\lambda }^{2}}}{h}\]
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