Q52 / 200 · NEET 2022 (Phase 2) · Chemistry
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?
$\begin{aligned}
& \left(h=6.626 \times 10^{-34} \mathrm{Js}, \mathrm{c}=3 \times 10^8 \mathrm{~ms}^{-1},\right. \\
& \mathrm{N}_{\mathrm{A}}=6.022 \times 10^{23} \mathrm{~mol}^{-1}
\end{aligned}$
- A$2.31 \times 10^6 \mathrm{~J} \mathrm{~mol}^{-1}$
- B$3.84 \times 10^4 \mathrm{~J} \mathrm{~mol}^{-1}$
- C$3.84 \times 10^{-19} \mathrm{~J} \mathrm{~mol}^{-1}$
- Correct answerD$2.31 \times 10^5 \mathrm{~J} \mathrm{~mol}^{-1}$
Solution
Since, the striking photon has energy equal to $h v \mathrm{v}$ and the minimum energy required to eject the electron is $h v_0$ (work function $\mathrm{W}_0$), the difference in energy $\left(h v-h v_0\right)$ is transferred as the kinetic energy of the photoelectron.
$h v=h v_0+\mathrm{KE}$
$h v_0=$ minimum energy required to remove electron)
Energy of a $300 \mathrm{~nm}$ photon is given by
$\begin{aligned}
& \Rightarrow \mathrm{E}=\frac{h c}{\lambda} \\
& =\frac{6.626 \times 10^{-34} \mathrm{Js} \times 3 \times 10^8 \mathrm{~ms}^{-1}}{300 \times 10^{-9} \mathrm{~m}} \\
& =6.626 \times 10^{-19} \mathrm{~J}
\end{aligned}$
Energy of one mole of photons
$\begin{aligned}
& =6.626 \times 10^{-19} \mathrm{~J} \times 6.022 \times 10^{23} \mathrm{~mol}^{-1} \\
& =3.99 \times 10^5 \mathrm{~J} \mathrm{~mol}^{-1}
\end{aligned}$
The minimum energy needed to remove one mole of electrons $=(3.99-1.68) \times 10^5 \mathrm{~J} \mathrm{~mol}^{-1}$ $=2.31 \times 10^5 \mathrm{~J} \mathrm{~mol}^{-1}$
Hence, the correct answer is option: 'D'.