Modern Physics Ques 100
- A beam of light has three wavelengths $4144 $ $\AA, 4972 $ $\AA$ and $6216 $ $\AA$ with a total intensity of $3.6 \times 10^{-3} Wm^{-2}$ equally distributed amongst the three wavelengths. The beam falls normally on an area $1.0 $ $cm^{2}$ of a clean metallic surface of work function $2.3$ $ eV$. Assume that there is no loss of light by reflection and that each energetically capable photon ejects one electron. Calculate the number of photoelectrons liberated in two seconds.
(1989, 8M)
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Answer:
Correct Answer: 100.$(1.1 \times 10^{12})$
Solution:
- Energy of photon having wavelength $4144 $ $\AA$,
$E _1=\frac{12375}{4144} $ $eV=2.99 $ $eV$
Similarly, $\quad E _2=\frac{12375}{4972}$ $ eV=2.49$ $ eV$ and
$ E _3=\frac{12375}{6216}$ $ eV=1.99$ $ eV $
Since, only $E _1$ and $E _2$ are greater than the work function $W=2.3$ $ eV$, only first two wavelengths are capable for ejecting photoelectrons. Given intensity is equally distributed in all wavelengths. Therefore, intensity corresponding to each wavelength is
$ \frac{3.6 \times 10^{-3}}{3}=1.2 \times 10^{-3} W / m^{2} $
Or energy incident per second in the given area ( $A=1.0 $ $cm^{2}=10^{-4}$ $ m^{2})$ is
$ \begin{aligned} \rho & =1.2 \times 10^{-3} \times 10^{-4} \\ & =1.2 \times 10^{-7} J / s \end{aligned} $
Let $n _1$ be the number of photons incident per unit time in the given area corresponding to first wavelength. Then
$ \begin{aligned} n _1 & =\frac{\rho}{E _1}=\frac{1.2 \times 10^{-7}}{2.99 \times 1.6 \times 10^{-19}} \\ & =2.5 \times 10^{11} \end{aligned} $
Similarly, $n _2=\frac{\rho}{E _2}=\frac{1.2 \times 10^{-7}}{2.49 \times 1.6 \times 10^{-19}}$
$ =3.0 \times 10^{11} $
Since, each energetically capable photon ejects electron, total number of photoelectrons liberated in $2$ $ s$.
$ \begin{aligned} & =2\left(n _1+n _2\right)=2(2.5+3.0) \times 10^{11} \\ & =1.1 \times 10^{12} \end{aligned} $
BETA
