A) \[\frac{hc}{e}\left[ \frac{1}{{{\lambda }_{3}}}+\frac{1}{2{{\lambda }_{2}}}-\frac{3}{2{{\lambda }_{1}}} \right]\]
B) \[\frac{hc}{e}\left[ \frac{1}{{{\lambda }_{3}}}+\frac{1}{{{\lambda }_{2}}}-\frac{3}{{{\lambda }_{1}}} \right]\]
C) \[\frac{hc}{e}\left[ \frac{1}{{{\lambda }_{3}}}+\frac{1}{2{{\lambda }_{2}}}-\frac{1}{{{\lambda }_{1}}} \right]\]
D) \[\frac{hc}{e}\left[ \frac{1}{{{\lambda }_{3}}}-\frac{1}{{{\lambda }_{2}}}-\frac{1}{{{\lambda }_{1}}} \right]\]
Correct Answer: A
Solution :
\[eV=\frac{hc}{{{\lambda }_{1}}}-{{\phi }_{0}}\] \[3eV=\frac{hc}{{{\lambda }_{2}}}-{{\phi }_{0}}\] \[eV'=\frac{hc}{{{\lambda }_{3}}}-{{\phi }_{0}}\] using these equations \[V'=\frac{hc}{e}\left( \frac{1}{{{\lambda }_{3}}}+\frac{1}{2{{\lambda }_{2}}}-\frac{3}{2{{\lambda }_{1}}} \right)\]You need to login to perform this action.
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