A) \[\Delta n=\frac{h}{2m\lambda }\]
B) \[\Delta n=\frac{h}{\lambda }\]
C) \[\left[ \frac{1}{{{v}_{0}}}-\frac{1}{v} \right]=\frac{m{{c}^{2}}}{n}\]
D) \[\lambda =\sqrt{\frac{h}{2m\Delta v}}\]
Correct Answer: D
Solution :
[d] According to the photoelectric equation, \[{{E}_{i}}=1E+KE\] or \[{{E}_{i}}\]= Threshold energy or work function \[+KE\] \[hv=h{{v}_{0}}+\frac{1}{2}m{{v}^{2}}\] \[\Rightarrow \,\,\,\frac{1}{2}m{{v}^{2}}=h(v-{{v}_{0}})'=h\Delta v\] ?..(i) We know that \[\lambda =\frac{h}{P}=\frac{h}{mv}\] \[\therefore \,\,\,v=\frac{h}{m\lambda }\] Substituting this value in equation (i), we get \[\frac{1}{2}m\frac{{{h}^{2}}}{{{m}^{2}}{{\lambda }^{2}}}=h\,\,\Delta v\] \[\frac{h}{2m{{\lambda }^{2}}}=\Delta v\] \[\therefore \,\,\,\lambda =\sqrt{\frac{h}{2m\Delta v}}\] Hence the answer is (4)
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