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When a beam of 10.6 \mathrm{eV} photons of intensity 2.0 \mathrm{~W} / \mathrm{m}^2 falls on a platinum surface of area 1.0 \times 10^{-4} \mathrm{~m}^2 and work function 5.6 \mathrm{eV} .  0 .53 \% of the incident photons eject photoelectrons. Find the number of photoelectrons emitted per.

Take 1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}.




 

Option: 1

2.5 \times 10^{12}


Option: 2

6.25 \times 10^{11}


Option: 3

4.5 \times 10^{11}


Option: 4

5 \times 10^{11}


Answers (1)

best_answer

Energy of incident photon.

\mathrm{E_1 =10.6 \mathrm{eV} }

\mathrm{ =10.6 \times 1.6 \times 10^{-19} \mathrm{~J} }

\mathrm{ =16.96 \times 10^{-19} \mathrm{~J}}

Energy incident per unit area per unit time (intensity) =2 \mathrm{~J}

\therefore Number of photons incident on unit area in unit time

=\frac{2}{16.96 \times 10^{-19}}

=1.18 \times 10^{18}
Therefore, number of photons incident per unit time on given are \left(1.0 \times 10^{-4} \mathrm{~m}^2\right)

=\left(1.18 \times 10^{18}\right)\left(1.0 \times 10^{-4}\right)

=1.18 \times 10^{18}
Therefore, number of photons incident per unit time on given area \left(1.0 \times 10^{-4} \mathrm{~m}^2\right)

=\left(1.18 \times 10^{18}\right)\left(1.0 \times 10^{-4}\right)

=1.18 \times 10^{14}

But only 0.53 \%of incident photons emit photoelectrons

\therefore Number of photoelectrons emitted per second \mathrm{(n)}

\mathrm{ n=\left(\frac{0.53}{100}\right)\left(1.18 \times 10^{14}\right) }

\mathrm{ n=6.25 \times 10^{11} }

 

Posted by

Rakesh

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