Atomic StructureEasy

Question

The threshold wavelength for ejection of electrons from a metal is 330 nm. The work function for the photoelectric emission from the metal is (h = 6.6 × 10–34 J-s)

Options

A.1.2 × 10–18 J
B.6.0 × 10–19 J
C.1.2 × 10–20 J
D.6.0 × 10–12 J

Solution

$\phi = \frac{hc}{\lambda} = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{330 \times 10^{- 9}} = 6 \times 10^{- 19}\text{ J}$

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