A mercury arc lamp provides $100 m W$ of UV radiation at a wavelength of $2480 Å$ (all other wavelengths having been absorbed by filters). The cathode of photoelectric device (a photo-tube) consists of potassium cathode and has an effective area of $4 c m^{2}$ . The anode is located at a distance of $1 m$ from radiation source. The work function $(ϕ_{0})$ for potassium is $2.25 e V$ .
Calculate the number of photons striking the cathode per second. Also calculate the saturation current, if the photo-conversion efficiency is $5 %$ (i.e., if each photon has a probability of $0.05$ of ejecting an electron).

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$8 \times 10^{12} s^{- 1}$ , 34 nA

$3 \times 10^{12} s^{- 1}$ , 30 nA

$4 \times 10^{12} s^{- 1}$ , 32 nA

$2 \times 10^{14} s^{- 1}$ , 36 nA

answer is B.

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Detailed Solution

Intensity at cathode $I = \frac{P}{4 {πr}^{2}} = \frac{100 \times 10^{- 3}}{4 π \times {(1)}^{2}} = \frac{0.1}{4 π} W / m^{2}$

Energy of each photon $E_{each photon} = \frac{h c}{λ} = \frac{12400}{2480} = 5 eV = 5 \times 1.6 \times 10^{- 19} J = 8 \times 10^{- 19} J$

Since, we know that the intensity

$\Rightarrow I = \frac{E_{total}}{A t} = (\frac{N}{t}) (\frac{E_{each photon}}{A})$

$\Rightarrow n = \frac{N}{t} = \frac{l A}{E_{each photon}}$

So, at the cathode $A = 4 \times 10^{- 4} m^{2}$ , the number of photons striking per second is

$n = \frac{I A}{E_{cach photon}} = (\frac{0.1}{4 π}) \frac{4 \times 10^{- 4}}{8 \times 10^{- 19}}$

$\Rightarrow n = 4 \times 10^{12} photons / s$

With an efficiency of $5 %$ , the photo-current

$I = (\frac{5}{100}) ne = (0.05) (4 \times 10^{12}) (1.6 \times 10^{- 19}) A$

$\Rightarrow I = 32 \times 10^{- 9} A = 32 nA$

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