A) \[1.76\times {{10}^{18}}\]photoelectrons/s
B) \[1.76\times {{10}^{14}}\]photoelectrons/s
C) \[8.8\times {{10}^{11}}\]photoelectrons/s
D) The value of work function is required to complete the value of emitted photoelectrons/s
Correct Answer: C
Solution :
Energy of photon, \[E=\frac{hc}{\lambda }\] \[=\frac{1242}{350}eV=3.55\,eV\] \[=5.68\times {{10}^{-19}}J\] Let n photons, per unit area per unit time are reaching the potassium surface, then \[n=\frac{1.00}{5.68\times {{10}^{-19}}}\] \[=1.76\times {{10}^{18}}\] So, number of photons received by potassium surface per unit time is, \[n\times \] Area of potassium surface \[=1.76\times {{10}^{18}}\times 1\times {{10}^{-4}}\] \[=1.76\times {{10}^{14}}\] Required number of photoelectrons emitted per unit time \[=1.76\times {{10}^{14}}\times \frac{0.5}{100}\] \[=8.8\times {{10}^{11}}\]
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