When electromagnetic radiation of wavelength 300 nm falls on the surface of a metal, electrons are emitted with the kinetic energy of 1.68 105 J mol1. What is the minimum energy needed to remove an electron from the metal?
(h = 6.626 1034 Js, c = 3 108 ms1, NA = 6.022 1023 mol1)
- A2.31 105 J mol1
- B2.31 106 J mol1
- C3.84 104 J mol1
- D3.84 1019 J mol1
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Correct answer: A
Energy of one photon $$ = {{hc} \over \lambda }$$ ($\lambda$ = 300 nm)
For one mole photons, $$E = {{hc} \over \lambda } \times {N_A}$$
$$E = {{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8} \times 6.023 \times {{10}^{23}}} \over {300 \times {{10}^{ - 9}}}}$$
$$E = 3.99 \times {10^5}$$ J mol$-$1
Kinetic energy $ = 1.68 \times {10^5}$ J mol$-$1
${W_0} = E - K.E.$
$$ = 3.99 \times {10^5} - 1.68 \times {10^5}$$
$ = 2.31 \times {10^5}$ J mol$-$1
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