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 $A^{\circ }$ and t is in second. The power incident on metal sheet is constant at 100 W. This signal is switched on and off for time intervals of 2 minutes and 1 minute respectively. Each time the signal is switched on, the $\lambda$ start from initial value of $3000A^{\circ }$. The metal plate is grounded and electron clouding is negligible. The efficiency of photoemission is 1% $(hc=12400eV{A}^{\circ }).$

The time after which photomission will stop is

• A. 79s
• B. 80s
• C. 81s
• D. 78s

Answer 1

The correct option is B 80s

Photo emission will stop if$\lambda ={\lambda }_0=\frac{hc}{\varphi }$

So, $3000+40t_0=\frac{12400}{2}$
which gives $t_0=80s$