Question

The work function of caesium metal is 2.14 eV. When the light of frequency $6\times {10}^{14}Hz$ is incident on the metal surface, photoemission of electrons occurs. Calculate:

(a) the maximum kinetic energy of the emitted electrons,

(b) stopping potential,

(c) the maximum speed of the emitted photoelectrons?

Answer 1

(a). Work function, ${\varphi }_o=2.14eV$

Frequency of light, $\nu =6\times {10}^{14}Hz$

Maximum K.E. is given by the photoelectric effect.

$K=h\nu -{\varphi }_o$

$K=\left(\frac{6.626\times {10}^{-34}\times 6\times {10}^{14}}{1.6\times {10}^{-19}}\right)-2.14=0.345eV$

(b). For stopping potential, $V_o$, we can write the equation for kinetic energy as,

$K=e{V}_o\Rightarrow V_o=\frac{K}{e}$

$=\frac{0.345\times 1.6\times {10}^{-19}}{1.6\times {10}^{-19}}=0.345V$

(c). The maximum speed of the emitted photoelectrons is $v$.

Hence, the relation for kinetic energy can be written as,

$K=\frac{1}{2}m{v}^2$

$m=9.1\times {10}^{-31}kg$ is the mass of electron.

$\Rightarrow v^2=2K/m$

$\Rightarrow v=3.323\times {10}^{5}m/s=332.3km/s$