Question

A monochromatic point source S radiating wavelength $6000 \overset{°}{A}$ with power $2$ watt, an aperture $A$ of diameter $0.1 m &$ a large screen $S C$ are placed as shown in figure. A photoemissive detector D of surface area $0.5 {cm}^{2}$ is placed at the centre of the screen. The efficiency of the detector for the photoelectron generation per incident photon is $0.9 .$

(i) Calculate the photon flux density at the centre of the screen and the photo current in the detector.

(ii) If a concave lens $L$ of focal length $0.6 m$ is inserted in the aperture as shown, find the new values of photon flux density & photocurrent. Assume a uniform average transmission of $80 %$ for the lens.

(iii) If the work-function of the photoemissive surface is $1 eV$ , calculate the values of the stopping potential in the two cases (without & with the lens in the aperture).

Accepted Answer

$(i)$ Power of source $= 2 J / s e c$

Energy of $1$ photon $= \frac{h c}{λ} = \frac{12400}{6000} = 2.067 eV$

No. of photons $L$

$= \frac{2 \times 10^{19}}{2.067 \times 16} = 6.0474 \times 10^{18}$

No. of photons striking on sphere of $0.6 m$ .

$= \frac{6.0474 \times 10^{18}}{4 π (0.6)^{2}} / m^{2}$

No. of photons passing through aperture

$= \frac{6.0474 \times 10^{18}}{4 π (0.6)^{2}} \times π {(\frac{0.1}{2})}^{2} = \frac{4.2 \times 10^{6}}{4} p h o t o n s$

No. of photons incident / area on screen (assuming aperture to act as a secondary point source)

$= \frac{4.2 \times 10^{16}}{4 \times 4 π (5.4)^{2}} = p h o t o n f l u x$

$= \frac{1.1462 \times 10^{14} photons}{m^{2} \sec}$

No. of photons incident on detector / sec

$= 1.1462 \times 10^{14} \times \frac{0.50}{10000}$

Photo current $= η \dot{N} e = 2.063 \times 10^{- 10} A$

Hence, photon flux density $1 {.1462\times10}^{14} \frac{Photons}{m^{2} - s}$ and photon current is $2.063 \times 10^{- 10} A$ .

$(ii)$ When the lens of $f = - 0.6 m$ is used,

When the lens of $f = - 0.6 m$ is used,

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v} = \frac{- 1}{0.3} = v = - 0.3 m$

Thus image will be at $0.3 m$ from the lens in the direction opposite to the screen.

Distance between screen $&$ image $= 5.7 m$

No. of photons striking lens

$=$ No. of photons striking the aperture

$= 1.05 \times 10^{16} p h o t o n s .$

Photons transmitted through the lens

$= 0 {.8\times1.05\times10}^{16} = \frac{0.84 \times 10^{16} p h o t o n}{s}$

This new situation, A point source emitting

$0.84 \times 10^{16}$ photons/ $\sec$ of $λ = 6000 \overset{°}{A}$

is kept at $5.7 m$ away from the screen.

Thus photons striking / area of screen

$= \frac{0.84 \times 10^{16}}{4 π (5.7)^{2}} = \frac{p h o t o n}{s e c} = 2.0574 \times 10^{13}$

Electrons emitted $= 0.9 \times 2.0574 \times 10^{13} \times \frac{0.5}{10000}$

Current $= 1.4813 \times 10^{- 10} A$

Hence, new flux density $2 {.0574\times10}^{13} \frac{Photons}{m^{2} - s}$ and photocurrent is $1.4813 \times 10^{- 10} A$

$(iii) λ = 6000 Å E_{p} = \frac{hc}{λ \times 1.6 \times 10^{- 19}} eV = 2.067 eV ϕ = 1 eV (given)$

$K . E ._{\max} = E_{p} - ϕ = 2.067 - 1 = 1.067 eV V_{s} = \frac{K . E ._{\max}}{e} = 1.067 V$

Hence, required stopping potential is $1.067 volt .$

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