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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