A photocell is operating in saturation mode having photo-current equal to 4.8mA, when a radiation $3000 \overset{\circ}{A}$ of and 1 W is incident on the photocathode. For another radiation of $1650 \overset{\circ}{A}$ and 5mW, maximum speed of photoelectrons is double of that for $3000 \overset{\circ}{A}$ radiation. Assuming quantum efficiency of photoelectron generation per incident photon equal in both cases, the threshold wavelength of the cell in $(\sqrt{x} - 1) μ m$ . Find x.

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Detailed Solution

As per Einstein’s equation,
$∴ K E_{\max} = E_{photon} - ϕ \Rightarrow \frac{1}{2} m v^{2} = \frac{h c}{λ} - ϕ$
For $λ = λ_{1} = 3000 \overset{\circ}{A}$ , we get
$\frac{1}{2} m v^{2} = \frac{h c}{λ_{2}} - ϕ$ …..(i)
For $λ = λ_{1} = 3000 \overset{\circ}{A}$ , as speed of electrons is doubled, so we get
As speed of electron is doubled, so we get
$\frac{1}{2} m {(2 v)}^{2} = \frac{h c}{λ^{2}} - ϕ$ ……..()
Dividing Eq. (ii) by (i), we get
$4 (\frac{h c}{λ_{1}} - ϕ) = \frac{h c}{λ_{2}} - ϕ \Rightarrow 3 ϕ = \frac{4 h c}{λ_{1}} - \frac{h c}{λ_{2}} \Rightarrow \frac{h c}{λ_{0}} = \frac{h c}{3} [\frac{4}{λ_{1}} - \frac{1}{λ_{2}}] \Rightarrow λ_{0} = \frac{3 λ_{1} λ_{2}}{4 λ_{2} - λ_{1}}$