A) 1.5 V
B) 1.85 V
C) 1.95 V
D) 2.37 V
Correct Answer: A
Solution :
\[\lambda =100nm=1000\overset{\text{o}}{\mathop{\text{A}}}\,\] Energy corresponding to \[1000\overset{\text{o}}{\mathop{\text{A}}}\,=\frac{12375}{1000}eV\] \[=12.375\text{ }eV\] Now, \[7.7=\text{ }12.375-{{\phi }_{0}}\] or \[{{\phi }_{0}}=12.375-7.7=4.675eV\] In the second case, Energy corresponding to \[2000\overset{\text{o}}{\mathop{\text{A}}}\,\] \[=\frac{12375}{2000}=6.1875eV\] Now, \[4.7=6.1875-\phi _{0}^{}\] or \[\phi _{0}^{}=6.1875=4.7\] \[=1.4875\] \[\simeq 1.5V\]
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