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Two separate monochromatic light beams A and B of the same intensity (energy per unit area per unit time) are falling normally on a unit area of a metallic surface. Their wavelength are $λ_{A} and λ_{B}$ respectively. Assuming that all the incident light is used in ejecting the photoelectrons, the ratio of the number of photoelectrons from beam A to that from B is

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${(\frac{λ_{B}}{λ_{A}})}^{2}$

$(\frac{λ_{A}}{λ_{B}})$

$(\frac{λ_{B}}{λ_{A}})$

${(\frac{λ_{A}}{λ_{B}})}^{2}$

answer is A.

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Detailed Solution

The number of photo electron depends on the Number of photons.

Number of photons emitted per unit time per unit area
$= \frac{I}{hc / λ} = \frac{λ \cdot I}{hc} \propto λ$
Ratio of no. of photo electrons $= \frac{λ_{A}}{λ_{B}}$

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