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Q.

In a photoelectric effect experiment, the threshold wavelength of light is 380 nm. If the wavelength of incident light is 260 nm, the maximum kinetic energy of emitted electrons will be $(x + 0.5) eV . Then x =$ (Given E (in eV) $= \frac{1237}{λ (in nm)}$ )

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a

1

b

10

c

2.5

d

4

answer is C.

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Detailed Solution

Given $λ_{0} = 380 n m, λ_{1} = 260 n m$

$K_{\max} = h (υ_{i} - υ_{0})$

$= h (\frac{c}{λ_{i}} - \frac{c}{λ_{0}})$

$= h c (\frac{λ_{0} - λ_{i}}{λ_{0} λ_{i}})$

$= 1237 (\frac{380 - 260}{380 \times 260}) e V = 1.5 e V$

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