Question

The electric field associated with a light wave is given by $E=E_0\sin \left[\right(1.57\times {10}^7\left)\right(x-ct\left)\right]$(where, x and t are in metre and second). The stopping potential when its light is used in an experiment on photoelectric effect with the emitter having work function $\varphi =19$eV is?

• A. $0.6$eV
• B. $1.2$eV
• C. $1.8$eV
• D. $2.4$eV

Answer 1

Answer: (B)

$1.2$eV

Given, $E=E_0\sin (1.57\times {10}^7)(x-ct)$ ........(i)

$\therefore E=E_0\sin \frac{\omega }{C}(x-ct)$ .........(ii)

On comparing Eqs. (i) and (ii), we get

$\frac{\omega }{C}=1.57\times {10}^7$

$=\frac{\pi }{2}\times {10}^7$

$\lambda =4\times {10}^{-7}$m

$=400$nm

As, ${\varphi }_0=\frac{hc}{\lambda }-E_k$

$1.9=\frac{1242}{400}-E_k$

$E_k=(3.1-1.9)eV=1.2$eV.