A) \[\Delta n=\frac{h}{2m\lambda }\]
B) \[\Delta n=\frac{h}{\lambda }\]
C) \[\left[ \frac{1}{{{v}_{0}}}-\frac{1}{v} \right]=\frac{m{{c}^{2}}}{h}\]
D) \[\lambda =\sqrt{\frac{h}{2m\Delta n}}\]
Correct Answer: D
Solution :
[d] \[{{E}_{1}}=IE+KE\] Or \[{{E}_{1}}=\]threshold E (or) work function +KE \[\left( hv=h{{v}_{0}}+\frac{1}{2}m{{u}^{2}} \right)\] Or \[hn=h{{n}_{0}}+\frac{1}{2}m{{u}^{2}}\] \[\frac{1}{2}m{{u}^{2}}=h(n-{{n}_{0}})=h\Delta n\] \[\left( \lambda =\frac{h}{mu},\therefore u=\frac{h}{m\lambda } \right)\] Substitute the value of u I equation (i) \[\frac{1}{2}m.\frac{{{h}^{2}}}{{{m}^{2}}{{\lambda }^{2}}}=h\Delta n\] \[\frac{h}{2{{\lambda }^{2}}m}=\Delta n\] \[\therefore \]\[\lambda =\sqrt{\frac{h}{2m\Delta n}}\]
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