Question

A beam of light consists of five wavelengths $4000\stackrel{\circ }{A},4800\stackrel{\circ }{A},6000\stackrel{\circ }{A},7000\stackrel{\circ }{A}and7800\stackrel{\circ }{A}$. The light beam is falling normally over a metal surface of area ${10}^{-4}{m}^2$ with a work function of 1.9 eV. Intensity of light beam is $7.5\times {10}^{-3}W/m^{-2}$, which is equally divided among the constituent wavelengths. If there is no loss of light energy, number of photoelectrons emitted per second is

• A. $1.12\times {10}^{12}$
• B. $3.15\times {10}^{12}$
• C. $1.77\times {10}^{12}$
• D. $4.06\times {10}^{12}$

Answer 1

The correct option is A $1.12\times {10}^{12}$

$E_1=\frac{12400}{4000}=3.1eV$
$E_2=\frac{12400}{4800}=2.58eV$
$E_3=\frac{12400}{6000}=2.06eV$
$E_4=\frac{12400}{7000}=1.77eV$
$E_5=\frac{12400}{7800}=1.58eV$
Clearly $4^{th}$ and $5^{th}$ wavelengths are not emitting any electron. Now number of photoelectrons emitted per second
= Number of photons incident per second
$=\frac{l_1{A}_1}{E_1}+\frac{l_2{A}_2}{E_2}+\frac{l_3{A}_3}{E_3}$
$=lA(\frac{1}{E_1}+\frac{1}{E_2}+\frac{1}{E_3})$
$=\frac{7.5\times {10}^{-3}}{5}\times {10}^{-4}\frac{1}{1.6\times {10}^{-19}}\times (\frac{1}{3.1}+\frac{1}{2.58}+\frac{1}{2})$