A) only A
B) A and B
C) All the three metals
D) None of these
Correct Answer: B
Solution :
\[E=\frac{hc}{\lambda }=\frac{6.63\times {{10}^{-34}}\times 3\times {{10}^{8}}}{41\times {{10}^{-8}}}J\]\[=\frac{19.89\times {{10}^{-18}}}{41\times 1.6\times {{10}^{-19}}}eV=3.0\,eV\] For metals A and B, the work function is less than 3.0 eV, so photoelectrons from metals A and B will be emitted with radiation of \[4000\,\overset{o}{\mathop{\text{A}}}\,\].
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