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Q.

A surface irradiated with light of wavelength 480 nm gives out electrons with maximum velocity 2 v m/s, the cut off wavelength being 600 nm. The same surface would release electrons with maximum velocity 2 v m/s, if it is irradiated by light of wavelength

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a

300 nm

b

360nm

c

384 nm

d

400 nm

answer is A.

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Detailed Solution

${(\frac{1}{2} {mv}^{2})}_{max} = \frac{hc}{480 \times 10^{- 9}} - \frac{hc}{600 \times 10^{- 9}} when 4 {(\frac{1}{2} {mv}^{2})}_{max} = \frac{hc}{λ} - \frac{hc}{600 \times 10^{- 9}} \frac{hc}{λ} - \frac{hc}{600 \times 10^{- 9}} = 4 [\frac{hc}{480 \times 10^{- 9}} - \frac{hc}{600 \times 10^{- 9}}] Solving, we get λ = 300 nm .$

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