In a photoelectric effect experiment, the threshold wavelength of light is 380nm. If the wavelength of incident light is 260nm, the maximum kinetic energy of emitted electrons will be: (in joules)

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$2.4 \times 10^{- 19} J$

$4.8 \times 10^{- 19} J$

$7.2 \times 10^{- 19} J$

$24.16 \times 10^{- 19} J$

answer is A.

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Detailed Solution

$\begin{array}{l} K E_{max} = E - ϕ_{o} \\ K E_{max} = \frac{h c}{λ} - \frac{h c}{λ_{o}} = 1237 [\frac{1}{260} - \frac{1}{380}] \\ = \frac{1237 \times 120}{380 \times 260} = 1.5 eV = 2.4 \times 10^{- 19} J \end{array}$

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