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In an experimental study of photoelectric effect, wavelength of incident radiation is $λ$ and kinetic energy of fastest moving electron is 4 eV. In the same experimental set up, when wavelength of incident radiation is $λ / 2$ , the energy of fastest moving electron is k, then

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$k' < 6 e V$

$k' = 6 eV$

$k' = 8 e V$

$k' > 8 eV$

answer is D.

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Detailed Solution

$\frac{hc}{λ} = ϕ + k . . . . . (1) and \frac{hc}{λ / 2} = ϕ + k' . . . . . (2)$

$∴ k' - k = \frac{hc}{λ} \Rightarrow k' = 2 k + (\frac{hc}{λ} - k) = 2 k + ϕ \Rightarrow k' > 2 k \Rightarrow k' > 8 eV$

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