When electromagnetic radiation of wavelength $300 \mathrm{~nm}$ falls on the surface of a metal, electrons…

When electromagnetic radiation of wavelength $300 \mathrm{~nm}$ falls on the surface of a metal, electrons are emitted with the kinetic energy of $1.68 \times 10^5$ $\mathrm{J} \mathrm{mol}^{-1}$. What is the minimum energy needed to remove an electron from the metal? $\begin{aligned} & \left(h=6.626 \times 10^{-34} \mathrm{Js}, \mathrm{c}=3 \times 10^8 \mathrm{~ms}^{-1},\right. \\ & \mathrm{N}_{\mathrm{A}}=6.022 \times 10^{23} \mathrm{~mol}^{-1} \end{aligned}$
  1. $2.31 \times 10^6 \mathrm{~J} \mathrm{~mol}^{-1}$
  2. $3.84 \times 10^4 \mathrm{~J} \mathrm{~mol}^{-1}$
  3. $3.84 \times 10^{-19} \mathrm{~J} \mathrm{~mol}^{-1}$
  4. $2.31 \times 10^5 \mathrm{~J} \mathrm{~mol}^{-1}$

Solution

Energy of one photon \(=\frac{h c}{\lambda}(\lambda=300 \mathrm{~nm})\)
For one mole photons, \(E=\frac{h c}{\lambda} \times N_A\)
\(\begin{aligned}
& E=\frac{6.626 \times 10^{-34} \times 3 \times 10^8 \times 6.023 \times 10^{23}}{300 \times 10^{-9}} \\
& E=3.99 \times 10^5 \mathrm{~J} \mathrm{~mol}^{-1}
\end{aligned}\)
Kinetic energy \(=1.68 \times 10^5 \mathrm{~J} \mathrm{~mol}^{-1}\)
\(\begin{aligned}
& W_0=E-K . E \\
& =3.99 \times 10^5-1.68 \times 10^5 \\
& =2.31 \times 10^5 \mathrm{~J} \mathrm{~mol}^{-1}
\end{aligned}\)

Asked in: NEET 2022 (Phase 2)

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