Question

The electric field associated with a light wave is $E=E_{0}sin\left[1.57\times {\left(10\right)}^7\left(x-ct\right)\right]$ where $x$ is in metre and $t$ is in second. If this light is used to produce photoelectric emission from the surface of a metal of work function $\text{1.9}eV$ , then the stopping potential will be

• A. $1.2V$
• B. $1.5V$
• C. $1.75V$
• D. $1.9V$

Answer 1

Answer: (A)

From $E=E_{0}sin\left[1.57\times {\left(10\right)}^7\left(x-ct\right)\right]$ frequency of incident wave is,
$v=\frac{\left(1.57\times {\left(10\right)}^7\right)c}{2\pi }$
For stopping potential,
$e{V}_s=hv-\phi$
$V_s=\frac{\left(6.626\times {\left(10\right)}^{-34}\right)\left(1.57\times {\left(10\right)}^7\right)\left(3\times {\left(10\right)}^8\right)}{2\pi \left(1.6\times {\left(10\right)}^{-19}\right)}-1.9$
$V_s=1.2V$