Question

Find the number of photons emitted per second by a 25Watt source of monochromatic light of wavelength 6600 Angstrom. What is the photoelectric current assuming 3% efficiency for photoelectric effect?
$\begin{align}
  & (A)\dfrac{25}{3}\times {{10}^{19}},0.4A \\
 & (B)\dfrac{25}{4}\times {{10}^{19}},6.2A \\
 & (C)\dfrac{25}{2}\times {{10}^{19}},0.8A \\
 & (D)\dfrac{25}{2}\times {{10}^{19}},4A \\
\end{align}$

Answer 1

Hint: Since each photon carries an energy of $hv$, and the power of light source is known. Therefore, the number of photons emitted will be equal to, Power upon energy of each photon. And the photoelectric current can be calculated as a product of efficiency, number of photoelectrons and charge on each photoelectron.

Complete answer:
Let the power of light source be denoted by P. And let the number of photons to be emitted be ‘n’. Then, we can write that:
$\Rightarrow P=n\left( \dfrac{hc}{\lambda } \right)$
Where, the term $\left( \dfrac{hc}{\lambda } \right)$is the energy of each photon.
Here, it has been given to us in the question:
$\Rightarrow P=25Watt$
$\begin{align}
  & \Rightarrow \lambda =6600\overset{\text{o}}{\mathop{\text{A}}}\, \\
 & \Rightarrow \lambda =6600\times {{10}^{-10}}m \\
\end{align}$
Also, we know that:
$\begin{align}
  & \Rightarrow h=6.626\times {{10}^{-34}}{{m}^{2}}kg{{s}^{-1}} \\
 & \Rightarrow c=3\times {{10}^{8}}m{{s}^{-1}} \\
\end{align}$
Hence, the number of photons emitted can be calculated as:
$\Rightarrow n=\dfrac{P\lambda }{hc}$
Therefore putting the values of all the respective terms in the right-hand side of the above equation, we get:
$\begin{align}
  & \Rightarrow n=\dfrac{25\times 6600\times {{10}^{-10}}}{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}} \\
 & \therefore n=\dfrac{25}{3}\times {{10}^{19}} \\
\end{align}$
Now, the photoelectric current (say I) can be calculated as follows:
$\Rightarrow I=\eta ne$
Where the given terms are:
$\eta $ is the efficiency of the photoelectric phenomenon
n is the number of photoelectrons emitted
e is the charge on each photoelectron
These terms have the given values:
$\begin{align}
  & \Rightarrow \eta =3\% \\
 & \Rightarrow \eta =\dfrac{3}{100} \\
\end{align}$
$\begin{align}
  & \Rightarrow n=\dfrac{25}{3}\times {{10}^{19}} \\
 & \Rightarrow e=1.6\times {{10}^{-19}}C \\
\end{align}$
Therefore putting all these values in the photoelectric current equation, we get:
$\begin{align}
  & \Rightarrow I=\dfrac{3}{100}\times \dfrac{25}{3}\times {{10}^{19}}\times 1.6\times {{10}^{-19}} \\
 & \Rightarrow I=\dfrac{40}{100}A \\
 & \therefore I=0.4A \\
\end{align}$
Hence, the number of photons emitted per second by the given light source is $\dfrac{25}{3}\times {{10}^{19}}$ and the current corresponding to this photoelectric effect is calculated to be $0.4A$ .

Hence, option (A) is the correct option.

Note:
As long and lengthy, these problems may seem at first. They are still pretty easy to solve. We can see the data given in the problem and data used are highly manipulative, that is, the whole equation solves to give a simple value. So, we shouldn’t skip easy problems like these in examinations and proceed ahead to solve them as these can get us some extra marks.