A) \[6\lambda \]
B) \[4\lambda \]
C) \[4\lambda /3\]
D) \[8\lambda \]
Correct Answer: B
Solution :
Einstein?s photoelectric equation, \[\frac{hc}{\lambda }=e(3{{V}_{0}})+W\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......(i)\] \[\frac{hc}{2\lambda }=e({{V}_{0}})+W\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......(i)\] Solving these equations, we get; \[\frac{hc}{\lambda }=\frac{3hc}{2\lambda }+W-3W\] \[or\,\,\,\,\,\,2W=\frac{hc}{2\lambda }\,\,\,\,or\,\,\,\,\frac{hc}{{{\lambda }_{0}}}=\frac{hc}{4\lambda }\] So, \[{{\lambda }_{0}}\,=\,\,4\lambda \]You need to login to perform this action.
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