First slide

Electromagnetic radiation and spectra

Question

The work function for caesium atom is 2 eV, the threshold wavelength in nm is

Moderate

A

654
B

754
C

854
D

954

Solution

$work function W_{0} = 2 eV = 2 \times 1.6 \times 10^{- 19} J = 3.2 \times 10^{- 19} J W_{0} = {hv}_{0} = \frac{hc}{λ_{0}} ∴ λ_{0} = \frac{hc}{W_{0}} = \frac{6 .626 \times 10^{- 34} Js \times 3 \times 10^{8} {ms}^{- 1}}{3.2 \times 10^{- 19} J} = 6 .5388 \times 10^{- 7} m = 653 .88 nm$

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