Question: 82: When electromagnetic radiation of wavelength 300 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?
( \mathrm{h}=6.626 \times 10^{-34} \mathrm{Js}, \mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1},
\mathrm{N}_{\mathrm{A}}=6.022 \times 10^{23} \mathrm{~mol}^{-1} )
(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}
Answer: Option (4)
Explanation:
This question is based on Einstein’s photoelectric equation.
According to the photoelectric effect, the energy of incident radiation is given by
E = \frac{hc}{\lambda}Substituting the given values,
E = \frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{300 \times 10^{-9}} E = 6.626 \times 10^{-19} \mathrm{~J \ per \ photon}Energy per mole of photons is obtained by multiplying with Avogadro number:
E_{\text{mole}} = 6.626 \times 10^{-19} \times 6.022 \times 10^{23} E_{\text{mole}} = 3.99 \times 10^{5} \mathrm{~J \ mol^{-1}}According to Einstein’s equation:
E = W + KEwhere W is the minimum energy (work function) required to remove an electron
and KE is the kinetic energy of emitted electrons.
Given kinetic energy of electrons:
KE = 1.68 \times 10^{5} \mathrm{~J \ mol^{-1}}Therefore,
W = E - KE W = 3.99 \times 10^{5} - 1.68 \times 10^{5} W = 2.31 \times 10^{5} \mathrm{~J \ mol^{-1}}Hence, the minimum energy needed to remove an electron from the metal is
2.31 \times 10^{5} \mathrm{~J \ mol^{-1}}.