AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

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).

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

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

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

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

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

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

Watch 3-min video & get full concept clarity

courses

No courses found

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!