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A photon of wavelength $1240Å$ is allowed to fall on a metal surface whose work function is 5 eV. Then kinetic energy of the photo electron emitted

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5 eV

10 eV

2 eV

6 eV

answer is A.

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Detailed Solution

$ϕ = \frac{hc}{λ} - K . E_{\max} 5 eV = \frac{6.634 \times 10^{- 34} \times 3 \times 10^{8}}{1240 \times 10^{- 10} \times 1.6 \times 10^{- 19}} - K . E_{\max} 5 eV = 10 eV - K . E_{\max} K . E_{\max} = 5 eV$

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