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 \[{{\lambda }_{th}}=\frac{hc}{W}=\frac{4e{{V}_{0}}\lambda }{e{{V}_{0}}}=4\lambda \]You need to login to perform this action.
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