Read the following passage and answer the questions.
A student studies a photocell. It is an evacuated glass tube with a quartz window. Inside the tube are an emitter plate C and a collector plate A. Plate C is coated with a metal that gives out electrons even in red light. A battery and a commutator let her make plate A positive or negative with respect to C. A microammeter reads the photocurrent.
In the first run, she shines a red laser pointer on plate C. The light has a wavelength of 620 nm, and the pointer gives out a power of 2.4 mW. She records the photocurrent as she changes the potential of A from positive to negative values. In the second run, she uses a red laser of the same wavelength but twice the power. In the third run, she uses a violet laser of wavelength 410 nm. She adjusts it so that the same number of photons reach plate C each second as in the first run.
(Take \(hc = 1240\ \mathrm{eV\,nm}\) and \(1\ \mathrm{eV} = 1.6\times10^{-19}\ \mathrm{J}\).)
The number of photons given out each second by the laser pointer in the first run is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(7.5\times10^{15}\)
The power of a beam is the energy it carries each second. Each photon carries the energy \( E=hc/\lambda \). So if \( N \) photons leave the pointer each second, the power is \( P=NE \).
Given:
\( \lambda=620\ \mathrm{nm} \)
\( P=2.4\ \mathrm{mW}=2.4\times10^{-3}\ \mathrm{W} \)
Energy of one photon:
\( \begin{aligned} E &= \frac{hc}{\lambda} \\ &= \frac{1240\ \mathrm{eV\,nm}}{620\ \mathrm{nm}} \\ &= 2.0\ \mathrm{eV} \end{aligned} \)
The power is in watts, so this energy is changed to joules:
\( \begin{aligned} E &= 2.0\times1.6\times10^{-19}\ \mathrm{J} \\ &= 3.2\times10^{-19}\ \mathrm{J} \end{aligned} \)
Number of photons per second:
\( \begin{aligned} N &= \frac{P}{E} \\ &= \frac{2.4\times10^{-3}}{3.2\times10^{-19}} \\ &= 7.5\times10^{15}\ \mathrm{s^{-1}} \end{aligned} \)
Therefore, the laser pointer gives out \( 7.5\times10^{15} \) photons each second.
NCERT (2026-27):
Particle Nature of Light: the Photon – “(iii) All photons of light of a particular frequency ν, or wavelength λ, have the same energy E (=hν = hc/λ) …”
Example 11.1 – “… If N is the number of photons emitted by the source per second, the power P transmitted in the beam equals N times the energy per photon E, so that P = N E.”