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. $1.2 \, V$
• B. $1.5 \, V$
• C. $1.75 \, V$
• D. $1.9 \, V$

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,
$eV_{s}=hv-\varphi$
$V_{s}=\frac{\left(6 .626 \times \left(10\right)^{- 34}\right) \left(1 .57 \times \left(10\right)^{7}\right) \left(3 \times \left(10\right)^{8}\right)}{2 \pi \left(1 .6 \times \left(10\right)^{- 19}\right)}-1.9$
$V_{s}=1.2V$