The photoelectric threshold wavelength of silver is $3250 \times 10^{- 10} m$ . The velocity of the electron ejected from a sliver surface by ultraviolet light of wavelength $2536 \times 10^{- 10} m$ is
(Given $h = 4.14 \times 10^{- 15} eVs$ and $c = 3 \times 10^{8} {ms}^{- 1}$ )

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$\approx$ 61 × 10³ ms^–1

$\approx$ 0.6 × 10⁵ ms^–1

$\approx$ 0.3 × 10⁶ ms^–1

$\approx$ 6 × 10⁵ ms^–1

answer is B.

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Detailed Solution

$\frac{hc}{λ} - \frac{hc}{λ_{0}} = \frac{1}{2} {mv}^{2}$

$\Rightarrow v^{2} = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{\frac{1}{2} \times 9 \times 10^{- 31}} [\frac{1}{2536} - \frac{1}{3250}] \times 10^{10}$

$\Rightarrow v = 6 \times 10^{5} m / s$

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