A) 5V
B) 2V
C) 0.754 V
D) 2.99V
Correct Answer: C
Solution :
[c] For transition from n=5 to n=4, \[hv=13.6\times 9\left[ \frac{1}{16}-\frac{1}{25} \right]=\frac{13.6\times 9\times 9}{16\times 25}=2.754eV\]For transition from n=4 to n=3, \[hv'=13.6\times 9\left[ \frac{1}{9}-\frac{1}{16} \right]=\frac{13.6\times 9\times 7}{9\times 16}=5.95eV\] For transition n=4 to n=3, the frequency is high and hence wavelength is short. For photoelectric effect, \[hv'=-W=e{{v}_{0}}\] Where W= work function \[5.95\times 1.6\times {{10}^{-19}}-W=1.6\times {{10}^{-19}}\times 3.95\] \[\Rightarrow W=2\times 1.6\times {{10}^{-19}}=2eV\] Again applying \[hv-W=eV{{'}_{0}}\] We get, \[2.754\times 1.6\times {{10}^{-19}}-2\times 1.6\times {{10}^{-19}}=\] \[1.6\times {{10}^{-19}}V{{'}_{0}}\]You need to login to perform this action.
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