A beam of light has three wavelengths `4144 Å`, and `6216 Å` with a total instensity of `3.6 xx 10^(-3) Wm^(-2)` equally distributed amongst the three wavelengths. The beam falls normally on an area `1.0 cm^2` of a clean metallic surface of work function 2.3 eV. Assume that there is no loss of light by reflection number of photoelectrons liberated in two seconds.

A beam of light has three wavelengths `4144 Å`, and `6216 Å` with a

1.	A beam of light has three wavelengths `4144 Å`, and `6216 Å` with a total instensity of `3.6 xx 10^(-3) Wm^(-2)` equally distributed amongst the three wavelengths. The beam falls normally on an area `1.0 cm^2` of a clean metallic surface of work function 2.3 eV. Assume that there is no loss of light by reflection number of photoelectrons liberated in two seconds.
Answer» Correct Answer - A::B Energy of photon having wavelength `4144 Å`, `E_1 = (12375)/(4144) eV` `=2.99 eV` Similarly, `E_2 = (12375)/(4972) eV` = 2.49 eV and `E_3 (12375)/(6216) eV` =1.99 eV Since, only `E_1` and `E_2` are greater than the work function `W = 2.3 ev`, only first two wavelengths are capable for ejecting photoelectrons. Given intensity is equally distributed in all wavelengths. Therefore, instensity corresponding to each wavelength is `(3.6xx10^(-3))/(3) = 1.2 xx10^(-3) W/m^2` Or energy incident per second in the given area (A = 1.0 cm^2 = 10^(-4) m^2)` is `P=1.2xx10^(-3)xx10^(-4)` `= 1.2 xx10^(-7) J/s` Let `n_1` be the number of photons incident per unit time in the ginven area corresponding to first wavelength. Then, `n_1 = (p)/(E_1)` `= (1.2xx10^(-7)/(2.99xx1.6xx10^(-19)` `=2.5xx10^11` Similarly, `n_2=(P)/(E_2)` `=1.2xx10^(-7)/(2.49xx1.6xx10^(-19)` `=3.0xx10^11` Since each energetically capable photon ejects one electron, total number of photoelectrons liberated in 2 s.

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