Modern Physics Ques 296
- When a beam of $10.6 $ $eV$ photons of intensity $2.0$ $ W / m^{2}$ falls on a platinum surface of area $1.0 \times 10^{-4} $ $m^{2}$ and work function $5.6$ $ eV .0 .53 \%$ of the incident photons eject photoelectrons. Find the number of photoelectrons emitted per second and their minimum and maximum energies (in $eV$ ). Take $1$ $ eV$ $=1.6 \times 10^{-19} J$.
$(2000,4 M)$
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Answer:
Correct Answer: 296.$(6.25 \times 10^{11}$, zero, $5.0 $ $eV)$
Solution:
Formula:
- Energy of incident photons,
$ \begin{aligned} E _i & =10.6 eV=10.6 \times 1.6 \times 10^{-19} J \\ & =16.96 \times 10^{-19} J \end{aligned} $
Energy incident per unit area per unit time (intensity) $=2 $ $ 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 area $\left(1.0 \times 10^{-4} m^{2}\right)$
$ =\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 (n)
$ \begin{aligned} n & =(\frac{0.53}{100})\left(1.18 \times 10^{14}\right) \\ n & =6.25 \times 10^{11} \\ K _{\min } & =0 \\ K _{\max } & =E _i-\text { work function } \\ & =(10.6-5.6) eV=5.0 eV \end{aligned} $
$ \begin{aligned} & \text { and } \quad K _{\max }=E _i \text {-work function } \\ & \therefore \quad K _{\max }=5.0 eV \\ & \text { and } \quad K _{\min }=0 \end{aligned} $
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