A) \[\frac{4\lambda }{3}\]
B) \[4\lambda \]
C) \[6\lambda \]
D) \[\frac{8\lambda }{3}\]
Correct Answer: B
Solution :
According to the question, \[ev=hc\left( \frac{1}{\lambda }-\frac{1}{{{\lambda }_{0}}} \right)\] ... (i) \[\frac{ev}{3}=hc\left( \frac{1}{2\lambda }-\frac{1}{{{\lambda }_{0}}} \right)\] ? (ii) Dividing Eq (i) by Eq (ii), we get \[3=\frac{\left( \frac{1}{\lambda }-\frac{1}{{{\lambda }_{0}}} \right)}{\left( \frac{1}{2\lambda }-\frac{1}{{{\lambda }_{0}}} \right)}\] or \[3\left( \frac{1}{2\lambda }-\frac{1}{{{\lambda }_{0}}} \right)=\frac{1}{\lambda }-\frac{1}{{{\lambda }_{0}}}\] \[\frac{3}{2\lambda }-\frac{1}{\lambda }=\frac{3}{{{\lambda }_{0}}}-\frac{1}{{{\lambda }_{0}}}\] \[\frac{1}{2\lambda }=\frac{2}{{{\lambda }_{0}}}\] Threshold wavelength for metallic surface \[{{\lambda }_{0}}=4\lambda \]
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