A photon of wavelength $3000 Å$ strikes a metal surface. The work function of the metal is $2.13…

A photon of wavelength $3000 Å$ strikes a metal surface. The work function of the metal is $2.13 \mathrm{eV}$. What is the kinetic energy of the emitted photoelectron? $\left(\mathrm{h}=6.626 \times 10^{-34} \mathrm{Js}\right)$
  1. $4.0 \mathrm{eV}$
  2. $3.0 \mathrm{eV}$
  3. $2.0 \mathrm{eV}$
  4. $1.0 \mathrm{eV}$

Solution

$\lambda=3000 Å=3 \times 10^{-7} \mathrm{~m}, \phi=2.13 \mathrm{eV}$ $\begin{aligned} & \text { K.E. }=h v-\mathrm{hv}_0=\mathrm{hv}-\phi=\frac{\mathrm{hc}}{\lambda}-\phi \\ & =\frac{\left(6.626 \times 10^{-34}-\frac{34}{2} \times 10^8\right)}{3 \times 10^{-7}}-\left(2.13 \times 1.6 \times 10^{-19}\right) \\ & =\frac{1.98 \times 10^{-25}}{3 \times 10^{-7}}-\left(3.408 \times 10^{-19}\right) \\ & =\left(6.60 \times 10^{-19}\right)-\left(3.408 \times 10^{-19}\right) \\ & =3.192 \times 10^{-19} \mathrm{~J} \\ & =2.0 \mathrm{eV}\end{aligned}$

Asked in: AP EAMCET 2023 (17 May Shift 2)

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