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  • A mono-chromatic point source S radiating wavelength  with power <img alt="\2 \

A mono-chromatic point source S radiating wavelength \mathrm{6000 \AA} with power \2 \mathrm{w}_1 an aperture A of diameter 0.1 metre and a large screen Sc are placed as shown in figure. A photo emission detector D of surface Area 0.5 \mathrm{~cm}^2 is placed at the centre of the screen. The efficiency of the detector for the photo-electro generation per incident photon is 0.9

Question:

Photo current in the detector

Option: 1

0.5 A


Option: 2

2.54 \times 10^{-12} \mathrm{~A}


Option: 3

9.65 \times 10^{-8} \mathrm{~A}


Option: 4

None


Answers (1)

best_answer

\mathrm{\text { Energy of one photon }=\frac{12400}{\lambda}=\frac{12400}{6000}=\frac{12.4}{6} \mathrm{eV}}

\mathrm{=\frac{12.4}{6} \times 1.6 \times 10^{-19} \text { Joule }=3.3 \times 10^{-19} \mathrm{~J}}

\mathrm{\text { Number of photon emitted per sec }=\frac{\text { Power }}{\text { Energy of onephoton }}=\frac{2}{3.3 \times 10^{-19}}=6.06\times 10^{18} / \mathrm{sec}}

Since aperture is large Photon flux at the centre of screen

\mathrm{=\frac{6.06 \times 10^{18}}{4 \pi(6)^2} \times\left(0.5 \times 10^{-4}\right)=0.0067 \times 10^{14}=6.7 \times 10^{11}}

\mathrm{\begin{gathered} \text { Photo current }=\text { efficiency } \times \text { photon flux } \times 1.6 \times 10^{-19} \\ =0.9 \times 6.7 \times 10^{11} \times 1.6 \times 10^{-19}=9.648 \times 10^{-8} \Lambda \end{gathered}}

Posted by

rishi.raj

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