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.