Question

The work function of caesium metal is $2.14 eV$. When light of frequency $6×1014Hz$ is incident on the metal surface,photoemission of electrons occurs. What is the
(a)maximum kinetic energy of the emitted electrons,
(b)Stopping potential, and
(c)maximum speed of the emitted photoelectrons?

Answer 1

Work function of caesium metal, $ϕº=2.14eV$
Frequency of light,$v=6.0×10^{14}HZ$
(a)The maximum kinetic energy is given by the photoelectric effect as:
$K=hv-ϕ_0$
Where,
h=Planck’s constant$=6.626×10^{-34}Js$
$∴K=\frac{6.626×10^{34}×6 \times 10^{14}}{1.6×10^{-19}}-2.14$
$=2.485-2.140=0.345eV$
Hence,the maximum kinetic energy of the emitted electrons is 0.345 eV.
(b)For stopping potential $V_0$,we can write the equation for kinetic energy as:
$K=eV_0$
$∴V_0=\frac{K}{e}$
$=\frac{0.345×1.6×10^{-19}}{1.6×10^{-19}}=0.345V$
Hence,the stopping potential of the material is 0.345 V.
(c)Maximum speed of the emitted photoelectrons= v
Hence,the relation for kinetic energy can be written as:
$K=\frac{1}{2}mv^2$
Where,
m=Mass of electron$=9.1×10^{-31}kg$
$v^2=\frac{2K}{m}$
$=\frac{2×0.345×1.6×10^{-19}}{9.1×10^{-31}}=0.1104×10^{12}$
$∴v=3.323×10^5m/s=332.3km/s$
Hence,the maximum speed of the emitted photoelectrons is 332.3 km/s.