Question

A light of wavelength $\lambda$ is incident on a metal sheet of work function $\varphi$ = 2eV. The wavelength $\lambda$ varies with time as$\lambda =3000+40t,$ where $\lambda$ is in $\mathring{A}$ and t is in second. The power incident on metal sheet is constant at $100W$. This signal is switched on and off for time intervals of $2mins$ and $1min$ respectively. Each time the signal is switched on, the $\lambda$ start from initial value of $3000$. The metal plate is grounded and electron clouding is negligible. The efficiency of photo-emission is $1\%$. $(hc=12400eV)$
The total number of photoelectrons ejected in one hour is

• A. $3.71\times {10}^{20}$
• B. $3.71\times {10}^{21}$
• C. $3.71\times {10}^{22}$
• D. $3.71\times {10}^{23}$

Answer 1

Answer: (B)

$3.71\times {10}^{21}$

Number of photons incident at any time t$(0\le t\le 80)$ in interval dt.
$d{n}_p=\frac{Pdt}{\left(\frac{hC}{\lambda }\right)}{n}_p={\int }_{t=0}^{t=80}\frac{\lambda Pdt}{hC}(P=100W)$
Number of photons emitted in one hour
$=\left(\frac{60min}{3min}\right)\times n_p$
Number of photoelectrons emitted in one hour -
$=\left(20n_p\right)\times 0.01=3.71\times {10}^{21}$

So, the answer is option (B)