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 \[\left( \frac{1}{2\lambda }-\frac{1}{{{\lambda }_{0}}} \right)=\frac{1}{\lambda }-\frac{1}{\lambda {{ & }_{0}}}\] \[\frac{3}{2\lambda }-\frac{1}{\lambda }=\frac{3}{{{\lambda }_{0}}}-\frac{1}{{{\lambda }_{0}}}\] \[\frac{1}{2\lambda }=\frac{2}{{{\lambda }_{0}}}\] Threshold wavelength for metallic surface \[{{\lambda }_{0}}=4\pi \]
You need to login to perform this action.
You will be redirected in
3 sec
