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 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 $2.25 e V$ .
According to classical theory, the radiation from the arc spreads out uniformly in space as spherical wave. Calculate the time of exposure of the metal to the radiation so that a potassium atom (radius 2 Å) in the anode accumulates sufficient energy to eject a photo-electron.

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360s

320s

310s

330s

answer is D.

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Detailed Solution

(a) The energy emitted per second per unit area (i.e. intensity) of the UV lamp at a distance of one metre is

$I = \frac{P}{4 π r^{2}} = \frac{100 \times 10^{- 3}}{4 π (1)^{2}} = \frac{0.1}{4 π} W m^{- 2}$

The cross-sectional area of atom i.e. effective area of the atom exposed to radiation is $(A_{eff} = π r^{2})$

$\Rightarrow A_{e f f} = π {(2 \times 10^{- 10} m)}^{2} = 4 π \times 10^{- 20} m^{2}$

Energy required to eject photo-electron from the metal surface is

$ϕ_{0} = (2.25) (1.6 \times 10^{- 19}) J = 3.6 \times 10^{- 19} J$

Since, $ϕ_{0} = I A t$ , hence the exposure time is given by

$t = \frac{ϕ_{0}}{l A_{e f f}} = \frac{3.6 \times 10^{- 19}}{(\frac{0.1}{4 π}) (4 π \times 10^{- 20})} = 360 s$

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