The work function of sodium metal is 4.41×10−19J. If photos 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=3×108 m/s____×10−21J

Solution:

$w_{0} = 4 .41 \times 10^{- 19} J λ = 300 nm, K . E = ? h = 6 .63 \times 10^{- 34} Js, w_{0} = hυ - K . E . 4 .41 \times 10^{- 19} J = \frac{6 .63 \times 10^{- 34} \times 3 \times 10^{8}}{300 \times 10^{- 9}} - K . E 4 .41 \times 10^{- 19} J = \frac{19 .89 \times 10^{- 26}}{300 \times 10^{- 9}} - K . E 4 .41 \times 10^{- 19} J = 6 .63 \times 10^{19} - K . E K . E = 2 .22 \times 10^{- 19} J = 222 \times 10^{- 21} J$

The work function of sodium metal is $4 .41 \times 10^{- 19} J .$ If photos of wavelength 300 nm are incident on the metal, the kinetic energy of the ejected electrons will be $(h = 6 .63 \times 10^{- 34} \times J s; C = 3 \times 10^{8} m / s)____\times 10^{- 21} J$

Solution:

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