A) 0.48 V
B) 2.0 V
C) 2.48 V
D) 0.72 V
Correct Answer: A
Solution :
\[\omega =6\times {{10}^{14}}\times 2\pi \] \[f=6\times {{10}^{14}}\] \[C=f\lambda \] \[\lambda =\frac{C}{f}=\frac{3\times {{10}^{8}}}{6\times {{10}^{14}}}=5000\overset{\text{o}}{\mathop{\text{A}}}\,\] energy of photon \[\Rightarrow \frac{12375}{5000}\] \[=2.475\text{ }eV\] from Einstein's equation \[K{{E}_{\max }}=E-\phi \] \[e{{V}_{s}}=E-\phi \] \[e{{V}_{s}}=2.475-2\] \[e{{V}_{s}}=0.475-2\] \[e{{V}_{s}}=0.475eV\] \[{{V}_{s}}=0.475V=0.48\]volt Option [a]
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