A) \[\frac{{{V}_{0}}}{2}\]
B) \[\,2\,{{V}_{0}}\]
C) \[{{V}_{0}}+\frac{hc}{2e{{\lambda }_{0}}}\]
D) \[{{V}_{0}}-\frac{hc}{2e{{\lambda }_{0}}}\]
Correct Answer: D
Solution :
As \[e{{V}_{0}}=\frac{hc}{{{\lambda }_{0}}}={{W}_{0}}\] ? (i) and \[eV'=\frac{hc}{2{{\lambda }_{0}}}-{{W}_{0}}\] ... (ii) Subtracting Eq. (i) from Eq. (ii) \[e({{V}_{0}}-V')\,\,=\,\,\frac{hc}{{{\lambda }_{0}}}\,\left( 1-\frac{1}{2} \right)\] \[=\,\,\,\frac{hc}{2{{\lambda }_{0}}}\] \[V'={{V}_{0}}-\frac{hc}{2e{{\lambda }_{0}}}\]You need to login to perform this action.
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