Question

Flat aluminum electrode is illuminated by light with a wavelength of $83nm$. What is the maximum distance (in mm) from the electrode surface will a photoelectron move, if outside the electrode there is a retarding electric field of $E=750V/m$. Photoelectric threshold for aluminum is 332 nm. (Take $hc=1245eV.nm$)

Answer 1

From Einstein's photo electric relation we can write,
$K{E}_{max}=\frac{hc}{\lambda }-\frac{hc}{{\lambda }_{Th}}$
given $hc=1245eV-nm$
${\lambda }_{TH}=332nmK{E}_{max}=1245[\frac{1}{83}-\frac{1}{332}]$
$=1245\left[\frac{4-1}{332}\right]$
$=1245\left[\frac{3}{332}\right]$
$=\frac{45}{4}eV$
from work-kinetic energy theorem work done $=\Delta KE$
$qV=\frac{45}{4}eV$
$eEX=\frac{45}{4}eV$
$X=\frac{45V}{4\times 750}\frac{m}{V}$
X = 0.015 m
X = 15 mm