The minimum energy required to eject an electron from the surface of a specific metal is 4.15 × 10^5 J mol^ -1 . If the metal surface is exposed to electromagnetic radiation of wavelength 200 nm , what will be the

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Quantrex Academy · Accepted Answer

Correct answer: C. 1.84 × 10^5 J mol^ -1

The energy of one mole of incident photons is given by

E = \dfrac{hcN_A}{\lambda}

Substituting the given values:

E = \dfrac{6.626 \times 10^{-34} \times 3 \times 10^8 \times 6.022 \times 10^{23}}{200 \times 10^{-9}}

E = \dfrac{19.878 \times 10^{-26} \times 6.022 \times 10^{23}}{2 \times 10^{-7}}

E = 9.939 \times 10^{-19} \times 6.022 \times 10^{23}

E = 59.85 \times 10^4 \text{ J mol}^{-1} = 5.985 \times 10^5 \text{ J mol}^{-1}

According to Einstein's photoelectric equation, the maximum kinetic energy is:

K_{\text{max}} = E - W

K_{\text{max}} = 5.985 \times 10^5 - 4.15 \times 10^5

K_{\text{max}} = 1.835 \times 10^5 \text{ J mol}^{-1}

Rounding off to three significant figures, we get

1.84 \times 10^5 \text{ J mol}^{-1}

.

Answer:

1.84 \times 10^5 \text{ J mol}^{-1}

The minimum energy required to eject an electron from the surface...

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