A) \[{{\lambda }_{0}}=\frac{2mc{{\lambda }^{2}}}{h}\]
B) \[{{\lambda }_{0}}=\frac{2h}{mc}\]
C) \[{{\lambda }_{0}}=\frac{2{{m}^{2}}{{c}^{2}}{{\lambda }^{3}}}{h}\]
D) \[{{\lambda }_{0}}=\lambda \]
Correct Answer: A
Solution :
[a] The cut off wavelength is given by \[{{\lambda }_{0}}=\frac{hc}{eV}\] ?.(i) According to de Broglie equation \[\lambda =\frac{h}{p}=\frac{h}{\sqrt{2meV}}\] \[\Rightarrow {{\lambda }^{2}}=\frac{{{h}^{2}}}{2meV}\Rightarrow V=\frac{{{h}^{2}}}{2me{{\lambda }^{2}}}\] ???(ii) From (i) and (ii), \[{{\lambda }_{0}}=\frac{hc\times 2me{{\lambda }^{2}}}{e{{h}^{2}}}=\frac{2mc{{\lambda }^{2}}}{h}\]
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