Answer
$162nm$
Work Step by Step
Given Wavelength of the incident light is $\lambda=221nm=221\times10^{-9}m=2.21\times10^{-7}m$
Energy of incident photon is given by
$E=hf=\frac{hc}{\lambda}$
putting $h=6.63\times10^{-34}J.s$, $\lambda=2.21\times10^{-7}m$,, $c=3\times10^{8}m/s$
$E=hf=\frac{6.63\times10^{-34}J.s\times3\times10^{8}m/s}{2.21\times10^{-7}m}=9.0\times10^{-19}J$.......equation(1)
Maximum kinetic energy of emitted electron is given as $KE_{max}=3.28\times10^{-19}J$
so from equation $hf=KE_{max} +W_{0}$
$W_{0}=hf-KE_{max}$
$W_{0}= 9.0\times10^{-19}J-3.28\times10^{-19}J=5.72\times10^{-19}J$
so work function of the metal is $W_{0}=5.72\times10^{-19}J$
If i want new maximum kinetic energy of electron to be doubled .
Suppose my new photon energy is $E_{new}$ with wavelength $\lambda_{new}$
new $KE_{max,new}=2\times KE_{max}=2\times3.28\times10^{-19}J$
$KE_{max,new}=6.56\times10^{-19}J$
$E_{new}=KE_{max,new}+W_{0}=6.56\times10^{-19}J+5.72\times10^{-19}J$
$E_{new}=12.28\times10^{-19}J$
$\lambda_{new}=\frac{hc}{E_{new}}$
putting $h=6.63\times10^{-34}J.s$, $E_{new}=12.28\times10^{-19}J$, $c=3\times10^{8}m/s$
$\lambda=\frac{6.63\times10^{-34}J.s\times3\times10^{8}m/s}{12.28\times10^{-19}J} =1.61970\times10^{-7}m=161.97\times10^{-9}m$
$\lambda=161.97nm\approx162nm$
