A light source, emitting three wavelengths $5000 Å$ , $6000 Å$ , and $7000 Å$ , has a total power of $10^{- 3} W$ and a beam diameter of $2 m m$ . The power density is distributed equally amongst the three wavelengths. The beam shines normally on a metallic surface of area of $10^{- 4} m^{2}$ and having a work function of $1.9 e V$ . Assuming that each photon liberates an electron, calculate the charge emitted per second from the metal surface.

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$10.28 m C$

$71.28 m C$

$67.28 m C$

$9.43 m C$

answer is B.

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Detailed Solution

$Since, E (in e V) = \frac{12375}{λ (in Å)}$

$\Rightarrow E_{1} = \frac{12375}{5000} = 2.475 e V$

$E_{2} = \frac{12375}{6000} = 2.063 e V and$

$E_{3} = \frac{12375}{7000} = 1.768 e V$

Since, $W$ is $1.9 e V$ , only the photons of energy $E_{1}$ and $E_{2}$ can emit photoelectrons.

Charge emitted per second is

$\frac{q}{t} = e (\frac{P}{3}) (\frac{A}{{πr}^{2}}) (\frac{λ_{1}}{12375 e} + \frac{λ_{2}}{12375 e}) = \frac{P A}{3 {πr}^{2}} (\frac{λ_{1} + λ_{2}}{12375})$

$\Rightarrow \frac{q}{t} = \frac{10^{- 3} \times 10^{- 4}}{3 π \times {(10^{- 3})}^{2}} \times (\frac{5000 + 6000}{12375}) = 9.43 \times 10^{- 3} C = 9.43 mC$