Question

The work function of sodium metal is $4.41\times {10}^{–19}J$. If photons of wavelength $300nm$ are incident on the metal, the kinetic energy of the ejected electrons will be ($h=6.63\times {10}^{–34}Js;c=3\times {10}^8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$) __________ .

Answer 1

Step 1: Given information

The work function of sodium metal $\left(\Phi \right)$= $4.41\times {10}^{–19}J$

Wavelength$\left(\lambda \right)$= $300nm=300\times {10}^{-9}m$

$h=6.63\times {10}^{–34}Js$

$c=3\times {10}^8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Step 2: Calculate the kinetic energy of the ejected electrons

We know, $K{E}_{max}=\left[\frac{hc}{\lambda }\right]–ɸ$

Therefore,

$K{E}_{max}=\left[\frac{6.63x{10}^{-34}x3x{10}^8}{300x{10}^{-9}}\right]–4.41\times {10}^{–19}$

$$\begin{gathered} \Rightarrow K{E}_{max}=6.63\times {10}^{–19}–4.41\times {10}^{–19} \\ \Rightarrow K{E}_{max}\Rightarrow =222\times {10}^{–21}J \end{gathered}$$