A) \[4\lambda \]
B) \[8\lambda \]
C) \[\frac{4}{3}\lambda \]
D) \[6\lambda \]
Correct Answer: A
Solution :
According to Einsteins photoelectric equation \[eV=hc\left[ \frac{1}{\lambda }-\frac{1}{{{\lambda }_{0}}} \right]\] Ist case \[3e{{V}_{s}}=hc\left[ \frac{1}{\lambda }-\frac{1}{{{\lambda }_{0}}} \right]\] ?(i) lInd case \[e{{V}_{s}}=hc\left[ \frac{1}{2\lambda }-\frac{1}{{{\lambda }_{0}}} \right]\] ...(ii) Dividing Eq. (i) by Eq. (ii), we get\[{{\lambda }_{0}}=4\lambda \]
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