Question

The work function of cesium metal is 2.14 eV. when light of frequency $6\times {10}^{14}Hz$ is incident on the metal surface, photoemission of electron occurs. find the (i) energy of incident photons (ii) maximum kinetic energy of photoelectrons.

Answer 1

Given, work function $\varphi =2.1eV$

$\nu =6\times {10}^{14}Hz$

Hence $E=h\nu$; where $h=$Planck's constant

$=6.63\times {10}^{-34}\times 6\times {10}^{14}$

$=39.78\times {10}^{-20}$ Joule

$=\frac{39.78\times {10}^{-20}eV}{1.6\times {10}^{-19}}$

$\{1eV=1.6\times {10}^{-10}Joules\}$

$1)\therefore E=24.865\times {10}^{-20+19}$

$=24.865\times {10}^{-1}$

$=2.4865$

$=2.5eV$

and $\varphi =2.1eV$

$2)$ Hence, $E=\varphi +(KE{)}_{max}$

$\Rightarrow (KE{)}_{max}=E-\varphi$

$=2.5-2.1$

$=0.4eV$