Question

Work function of a metal is 3eV. An electromagnetic radiation of 200 nm is incident on metal surface, calculate the maximum speed of photoelectrons ejected. (Take the value of h= $6.6\times {10}^{-34}Js)$

• A. $2.05\times {10}^{6}m/s$
• B. $1.75\times {10}^{6}m/s$
• C. $1.05\times {10}^{6}m/s$
• D. $2.20\times {10}^{6}m/s$

Answer 1

The correct option is C $1.05\times {10}^{6}m/s$

Energy of incident radiation $=h\nu =\frac{hc}{\lambda }$
$=\frac{6.6\times {10}^{-34}Js\times 3\times {10}^{8}m/s}{200\times {10}^{-9}m}=9.90\times {10}^{-19}J\dots .\left(1\right)$
Given value of work function = 3 eV
Converting 3 eV to Joules =$3\times 1.6\times {10}^{-19}J=4.80\times {10}^{-19}J$
Now we know from photoelectric equation,
Incident radiation energy = work function + kinetic energy
$h\nu =h{\nu }_0+\frac{1}{2}{m}_e{V}_{max}^2$
By substituting values,
$9.90\times {10}^{-19}J=4.80\times {10}^{-19}J+\frac{1}{2}{m}_e{V}_{max}^2\frac{1}{2}{m}_e{V}_{max}^2=5.10\times {10}^{-19}\frac{1}{2}\times 9.1\times {10}^{-31}kg{V}_{max}^2=5.10\times {10}^{-19}{V}_{max}=1.05\times {10}^{6}m/s$