Answer
Maximum kinetic energy of the electron which is ejected from the metal with work function $W_{0}=2.17eV$, will be $KE_{max}=1.26eV$
Work Step by Step
Given that Maximum kinetic energy of electron $KE_{max}=0.68eV$
work function of the metal $W_{0}=2.75eV$
suppose Energy of incident photon is $E=hf$
so from equation 29.3
$hf=KE_{max}+W_{0}$
putting the values we will get
$hf=0.68eV+2.75eV=3.43eV$
so Energy of incident photon $E=hf=3.43eV$
This same radiation incident upon another metal whose work function is $W_{0}=2.17eV$
so again from equation $hf=KE_{max}+W_{0}$
$KE_{max}=hf-W_{0}$
putting the values $W_{0}=2.17eV$, $E=hf=3.43eV$
we will get $KE_{max}=3.43eV-2.17eV=1.26eV$
