Modern Physics Ques 96
- Two metallic plates $A$ and $B$ each of area $5 \times 10^{-4} m^{2}$, are placed parallel to each other at separation of $1 $ $cm$. Plate $B$ carries a positive charge of $33.7 \times 10^{-12} C$. A monochromatic beam of light, with photons of energy $5 $ $eV$ each, starts falling on plate $A$ at $t=0$ so that $10^{16}$ photons fall on it per square metre per second. Assume that one photoelectron is emitted for every $10^{6}$ incident photons. Also assume that all the emitted photoelectrons are collected by plate $B$ and the work function of plate $A$ remains constant at the value $2 $ $eV$. Determine
$(2002,5 M)$
(a) the number of photoelectrons emitted up to $t=10 $ $s$,
(b) the magnitude of the electric field between the plates $A$ and $B$ at $t=10$ $ s$ and
(c) the kinetic energy of the most energetic photoelectrons emitted at $t=10 $ $s$ when it reaches plate $B$.
Neglect the time taken by the photoelectron to reach plate $B$.
(Take $\varepsilon _0=8.85 \times 10^{-12} C^{2} / N-m^{2}$ ).
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Answer:
Correct Answer: 96.(a) $5.0 \times 10^{7}\quad $ (b) $2 \times 10^{3} N / C \quad $ (c) $ 23$ $ eV$
Solution:
Formula:
- Area of plates $A=5 \times 10^{-4} m^{2}$
Distance between the plates $d=1 $ $cm=10^{-2} m$
(a) Number of photoelectrons emitted upto $t=10 $ $ s$ are
$ n=\frac{\text {( number of photons falling in unit area in unit time )} \times(\text { area } \times \text { time })}{10^{6}} $
$ =\frac{1}{10^{6}}\left[(10)^{16} \times\left(5 \times 10^{-4}\right) \times(10)\right]=5.0 \times 10^{7}$
(b) At time, $t=10$ $ s$
Charge on plate $A, q _A=+n e=\left(5.0 \times 10^{7}\right)\left(1.6 \times 10^{-19}\right)$
$ =8.0 \times 10^{-12} C $
and charge on plate $B$,
$ \begin{aligned} q _B & =\left(33.7 \times 10^{-12}-8.0 \times 10^{-12}\right) \\ & =25.7 \times 10^{-12} C \end{aligned} $
$\therefore$ Electric field between the plates, $E=\frac{\left(q _B-q _A\right)}{2 A \varepsilon _0}$
or $E=\frac{(25.7-8.0) \times 10^{-12}}{2 \times\left(5 \times 10^{-4}\right)\left(8.85 \times 10^{-12}\right)}=2 \times 10^{3} N / C$
(c) Energy of photoelectrons at plate $A$
$ =E-W=(5-2) eV=3 $ $eV $
Increase in energy of photoelectrons
$ \begin{aligned} & =(e E d) \text { joule }=(E d) eV \\ & =\left(2 \times 10^{3}\right)\left(10^{-2}\right) eV=20 eV \end{aligned} $
Energy of photoelectrons at plate $B$
$ =(20+3) eV=23$ $ eV $
BETA
