The work function of sodium metal is 4.41 × $10^{- 19}$ J. If photon of wavelength 300 nm are incident on the metal, the kinetic energy of the ejected electrons will be (h =6.63 × $10^{- 34}$ J s; c =3 × $10^{8}$ m/s) _____ × $10^{- 21}$ J.

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answer is 222.

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Detailed Solution

$KE = \frac{hc}{λ} - W$

$KE = \frac{6 .63 \times 10^{- 34} \times 3 \times 10^{8}}{3 \times 10^{- 7}} - 4 .41 \times 10^{- 19} KE = 6 .63 \times 10^{- 19} - 4 .41 \times 10^{- 19} KE = 2 .22 \times 10^{- 19} J or 222 \times 10^{- 21} J$

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