A) \[6\lambda \]
B) \[4\lambda \]
C) \[\frac{\lambda }{4}\]
D) \[\frac{\lambda }{6}\]
Correct Answer: B
Solution :
From photoelectric equation |
\[hv+W+e{{V}_{0}}\] |
(where, W = work function) |
So \[\frac{hc}{\lambda }=\,W+3e{{V}_{0}}\] (i) |
Also, \[\frac{hc}{2\lambda }=W+e{{V}_{0}}\] |
\[\Rightarrow \] \[\frac{hc}{\lambda }=2W+2e{{V}_{0}}\] |
Subtracting Eq. (i) from Eq. (ii), we get |
\[0=W-e{{V}_{0}}\] |
\[\Rightarrow \] \[W=e{{V}_{0}}\] |
From Eq. (i), |
\[\frac{hc}{\lambda }=e{{V}_{0}}+3e{{V}_{0}}=4e{{V}_{0}}\] |
The threshold wavelength is given by |
\[{{\lambda }_{th}}=\frac{hc}{W}=\frac{4e{{V}_{0}}\lambda }{e{{V}_{0}}}=4\lambda \] |
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