Discussion
Work function of Al is 4.2 eV. When light with E = 6.2 eV is incident on an Al surface, maximum kinetic energy of the emitted photo-electrons will be (1 eV = 1.6 × 10-19 J) :
Answer: C
The maximum kinetic energy is 3.2 × 10-19 J.
By Einstein's photoelectric equation, the photon energy goes into overcoming the work function, and the rest appears as the kinetic energy of the fastest electrons:
\[ \begin{aligned} K.E._{max} &= h\nu - W_{0} = 6.2 - 4.2 = 2.0\ \text{eV} \\ &= 2.0 \times 1.6 \times 10^{-19} \\ &= 3.2 \times 10^{-19}\ \text{J} \end{aligned} \]
2.0 × 10-19 J leaves the answer in electron-volts while writing joules, and 9.9 × 10-19 J and 6.7 × 10-19 J correspond to converting the photon energy or the work function instead of their difference.
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