What is the work function of the metal if the light of wavelength
Question:

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 $=1.6 \times 10^{-19} \mathrm{JeV}^{-1}$ )

  1. $0.9 \mathrm{eV}$

  2. $3.1 \mathrm{eV}$

  3. $2.1 \mathrm{eV}$

  4. $4.0 \mathrm{eV}$


Correct Option: , 3

Solution:

$E=h v=\frac{\mathrm{hc}}{\lambda}$

$E=\frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{4000 \times 10^{-10} \times 1.6 \times 10^{-19}}=3.1 \mathrm{eV}$

$\mathrm{KE}=\frac{1}{2} m v^{2}=\frac{1}{2} \times 9 \times 10^{-31} \times 36 \times 10^{10} \mathrm{~J}$

$=1.62 \times 10^{-19} \mathrm{~J}$

$=1 \mathrm{eV}$

According to photoelectric effect,

K.E. $=h v-h v_{0}$

$h v_{0}=h v-\mathrm{K} . \mathrm{E} .$

Work function $\left(W_{0}\right)=E-\mathrm{K} . \mathrm{E}$.

$=3.1-1=2.1 \mathrm{eV}$

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