A) none
B) A only
C) A and B only
D) all the three metals
Correct Answer: C
Solution :
Key Idea: That metal will emit photoelectrons which has work function lower than that obtained with the radiation of \[4100\,\overset{\text{o}}{\mathop{\text{A}}}\,.\] Work function for wavelength of \[4100\,\overset{\text{o}}{\mathop{\text{A}}}\,\] is \[W=\frac{hc}{\lambda }\] \[=\frac{6.62\times {{10}^{-34}}\times 3\times {{10}^{8}}}{4100\times {{10}^{-10}}}\] \[=4.8\times {{10}^{-19}}J\] \[=\frac{4.8\times {{10}^{-19}}}{1.6\times {{10}^{-19}}}eV=3\,eV\] Now, we have \[{{W}_{A}}=1.92\,eV,\] \[{{W}_{B}}=2.0\,eV,\] \[{{W}_{C}}=5\,eV\] Since, \[{{W}_{A}}<W\] and \[{{W}_{B}}<W,\] hence, A and B will emit photoelectrons.
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