A) \[1.18\times {{10}^{10}},2eV,5eV\]
B) \[1.18\times {{10}^{14}},0eV,5eV\]
C) \[2.18\times {{10}^{13}},0eV,5eV\]
D) \[3.11\times {{10}^{11}},1eV,5eV\]
Correct Answer: B
Solution :
[b] No. of photons/sec \[\text{=}\frac{\text{Energy incident on platinum surface per second}}{\text{Energy of one photon}}\]No. of photons incident per second \[=\frac{2\times 10\times {{10}^{-4}}}{10.6\times 1.6\times {{10}^{-19}}}=1.18\times {{10}^{14}}\] As 0.53% of incident photon can eject photoelectrons \[\therefore \]No. of photoelectrons ejected per second \[=1.18\times {{10}^{14}}\times \frac{0.53}{100}=6.25\times {{10}^{11}}\] As 0.53% of incident photon can eject photoelectrons Minimum energy \[=0eV,\] Maximum energy \[=\left( 10.6-5.6 \right)eV=5eV\]
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