What is the work function of the metal if the light of wavelength $4000 \AA$ generates photoelectrons of velocity $6 \times 10^{5} \mathrm{~ms}^{-1}$ from it?
(Mass of electron $=9 \times 10^{-31} \mathrm{~kg}$
Velocity of light $=3 \times 10^{8} \mathrm{~ms}^{-1}$
Planck's constant $=6.626 \times 10^{-34} \mathrm{Js}$
Charge of electron $=1.6 \times 10^{-19} \mathrm{JeV}^{-1}$ )
Correct Option: , 3
$E=h v=\frac{\mathrm{hc}}{\lambda}$
$E=\frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{4000 \times 10^{-10} \times 1.6 \times 10^{-19}}=3.1 \mathrm{eV}$
$\mathrm{KE}=\frac{1}{2} m v^{2}=\frac{1}{2} \times 9 \times 10^{-31} \times 36 \times 10^{10} \mathrm{~J}$
$=1.62 \times 10^{-19} \mathrm{~J}$
$=1 \mathrm{eV}$
According to photoelectric effect,
K.E. $=h v-h v_{0}$
$h v_{0}=h v-\mathrm{K} . \mathrm{E} .$
Work function $\left(W_{0}\right)=E-\mathrm{K} . \mathrm{E}$.
$=3.1-1=2.1 \mathrm{eV}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.