Photons of energy $10 \mathrm{eV}$ are incident on a photosensitive surface of threshold frequency $2 \times…

Photons of energy $10 \mathrm{eV}$ are incident on a photosensitive surface of threshold frequency $2 \times 10^{15} \mathrm{~Hz}$. The kinetic energy in $\mathrm{eV}$ of the photoelectrons emitted is [Planck's constant $\mathrm{h}=6.63 \times 10^{34}$ $\mathrm{Js}]$
  1. $8.29 \mathrm{eV}$
  2. $6.5 \mathrm{eV}$
  3. $4.2 \mathrm{eV}$
  4. $1.7 \mathrm{eV}$

Solution

$\begin{aligned} & \text { Energy of photons, } \mathrm{h} v=10 \mathrm{eV} \\ & \text { Threshold frequency }=2 \times 10^{15} \mathrm{~Hz} \\ & \therefore \text { Work function, } \mathrm{W}_0=\mathrm{h} v_0 \\ & =6.63 \times 10^{-34} \times 2 \times 10^{15} \\ & =13.26 \times 10^{-19} \mathrm{~J} \\ & =\frac{13.26 \times 10^{-19}}{1.6 \times 10^{-19}} \mathrm{eV}=8.3 \mathrm{eV} \\ & \mathrm{KE} \text { of electrons }=\mathrm{h} v-\mathrm{W}_0 \\ & =10-8.3=1.7 \mathrm{eV}\end{aligned}$

Asked in: MHT CET 2021 (22 Sep Shift 2)

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