A) \[4\lambda /3\]
B) \[4\lambda\]
C) \[6\lambda\]
D) \[8\lambda /3\]
Correct Answer: B
Solution :
According to the questions \[eV=hc\left( \frac{1}{\lambda }-\frac{1}{{{\lambda }_{0}}} \right)\] ?.(i) And \[\frac{eV}{3}=hc\left( \frac{1}{2\lambda }-\frac{1}{{{\lambda }_{0}}} \right)\] ?(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 \[3\left( \frac{1}{2\lambda }-\frac{1}{{{\lambda }_{0}}} \right)=\frac{1}{\lambda }-\frac{1}{{{\lambda }_{0}}}\] Or \[\frac{3}{2\lambda }-\frac{1}{\lambda }=\frac{3}{{{\lambda }_{0}}}-\frac{1}{{{\lambda }_{0}}}\] Or \[\frac{1}{2\lambda }=\frac{2}{{{\lambda }_{0}}}\] Threshold wavelength for metallic surface, \[{{\lambda }_{0}}=4\lambda\]
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