Particle Nature of Light–The Photon — NEET Physics PYQ MCQs with Solutions
Free NEET Physics PYQ Particle Nature of Light–The Photon MCQs with step-by-step solutions (13 questions). Part of Dual Nature of Radiation and Matter. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Particle Nature of Light–The Photon · medium · numerical
The number of photons per second on an average emitted by the source of monochromatic light of wavelength 600 nm, when it delivers the power of $3.3 \times 10^{-3}$ W will be ($h = 6.6 \times 10^{-34}$ J-s)
A. $10^{18}$
B. $10^{17}$
C. $10^{16}$ ✓ Correct
D. $10^{15}$
Solution: $P = \frac{nhc}{\lambda}$
$n = \frac{P\lambda}{hc} = \frac{3.3 \times 10^{-3} \times 600 \times 10^{-9}}{6.6 \times 10^{-34} \times 3 \times 10^8} = 10^{16}$ photons/s
Q2 — Particle Nature of Light–The Photon · medium · numerical
A 200 W sodium street lamp emits yellow light of wavelength 0.6 µm. Assuming it to be 25% efficient in converting electrical energy to light, the number of photons of yellow light it emits per second is
A. $1.5 \times 10^{20}$ ✓ Correct
B. $6 \times 10^{18}$
C. $62 \times 10^{20}$
D. $3 \times 10^{19}$
Solution: Useful power $= 25\%$ of 200 W $= 50$ W
$\frac{N}{t} = \frac{P\lambda}{hc} = \frac{50 \times 0.6 \times 10^{-6}}{6.6 \times 10^{-34} \times 3 \times 10^8} = 1.5 \times 10^{20}$ photons/s
Q3 — Particle Nature of Light–The Photon · medium · numerical
A source $S_1$ is producing $10^{15}$ photons/s of wavelength 5000 Å. Another source $S_2$ is producing $1.02 \times 10^{15}$ photons per second of wavelength 5100 Å. Then, (power of $S_2$)/(power of $S_1$) is equal to
A. 1.00 ✓ Correct
B. 1.02
C. 1.04
D. 0.98
Solution: $P = \frac{nhc}{\lambda}$
$\frac{P_2}{P_1} = \frac{n_2\lambda_1}{n_1\lambda_2} = \frac{1.02 \times 10^{15} \times 5000}{10^{15} \times 5100} = 1.00$
Q4 — Particle Nature of Light–The Photon · medium · numerical
Monochromatic light of wavelength 667 nm is produced by a helium neon laser. The power emitted is 9 mW. The number of photons arriving per second on the average at a target irradiated by this beam is
A. $9 \times 10^{17}$
B. $3 \times 10^{16}$ ✓ Correct
C. $9 \times 10^{15}$
D. $3 \times 10^{19}$
Solution: $N = \frac{P\lambda}{hc} = \frac{9 \times 10^{-3} \times 667 \times 10^{-9}}{6.6 \times 10^{-34} \times 3 \times 10^8} = 3 \times 10^{16}$ photons/s
Q5 — Particle Nature of Light–The Photon · medium · numerical
Monochromatic light of frequency $6.0 \times 10^{14}$ Hz is produced by a laser. The power emitted is $2 \times 10^{-3}$ W. The number of photons emitted, on the average, by the source per second is
A. $5 \times 10^{15}$ ✓ Correct
B. $5 \times 10^{16}$
C. $5 \times 10^{17}$
D. $5 \times 10^{14}$
Solution: Energy per photon: $E = h\nu = 6.6 \times 10^{-34} \times 6 \times 10^{14}$ J
$n = \frac{P}{E} = \frac{2 \times 10^{-3}}{6.6 \times 10^{-34} \times 6 \times 10^{14}} = 5 \times 10^{15}$
Q6 — Particle Nature of Light–The Photon · medium · numerical
The momentum of a photon of energy 1 MeV, in kg m/s, will be
A. $0.33 \times 10^6$
B. $7 \times 10^{-24}$
C. $10^{-22}$
D. $5 \times 10^{-22}$ ✓ Correct
Solution: For a photon, $p = \frac{E}{c}$
$p = \frac{1 \times 10^6 \times 1.6 \times 10^{-19}}{3 \times 10^8} = 5 \times 10^{-22}$ kg-m/s
Q7 — Particle Nature of Light–The Photon · medium · numerical
The 21 cm radiowave emitted by hydrogen in interstellar space is due to the interaction called the hyperfine interaction in atomic hydrogen. The energy of the emitted wave is nearly
A. $10^{-17}$ J
B. 1 J
C. $7 \times 10^{-6}$ J
D. $10^{-24}$ J ✓ Correct
Solution: $E = \frac{hc}{\lambda} = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{0.21} \approx 10^{-24}$ J
Q8 — Particle Nature of Light–The Photon · easy · theory
The nature of ions knocked out from hot surfaces is
A. protons
B. electrons ✓ Correct
C. neutrons
D. nuclei
Solution: Heating a metal surface causes thermionic emission — the particles knocked out are electrons.
Q9 — Particle Nature of Light–The Photon · easy · theory
Momentum of a photon of wavelength $\lambda$ is
A. $\frac{h}{\lambda}$ ✓ Correct
B. zero
C. $\frac{h\lambda}{c^2}$
D. $\frac{h\lambda}{c}$
Solution: A photon moves with speed c, so its momentum is
$p = \frac{h\nu}{c} = \frac{h}{\lambda}$
Q10 — Particle Nature of Light–The Photon · medium · numerical
The wavelength of a 1 keV photon is $1.24 \times 10^{-9}$ m. What is the frequency of 1 MeV photon?
A. $1.24 \times 10^{15}$ Hz
B. $2.4 \times 10^{20}$ Hz ✓ Correct
C. $1.24 \times 10^{18}$ Hz
D. $2.4 \times 10^{23}$ Hz
Solution: For the 1 keV photon: $\frac{hc}{\lambda} = 10^3$ eV gives $h = \frac{10^3\lambda}{c}$ (in eV units)
For 1 MeV: $h\nu = 10^6$ eV
$\nu = \frac{10^3 c}{\lambda} = \frac{10^3 \times 3 \times 10^8}{1.24 \times 10^{-9}} = 2.4 \times 10^{20}$ Hz
Q11 — Particle Nature of Light–The Photon · medium · numerical
The momentum of a photon of an electromagnetic radiation is $3.3 \times 10^{-29}$ kg-ms⁻¹. What is the frequency of the associated waves? ($h = 6.6 \times 10^{-34}$ J-s, $c = 3 \times 10^8$ ms⁻¹)
A. $1.5 \times 10^{13}$ Hz ✓ Correct
B. $7.5 \times 10^{12}$ Hz
C. $6.0 \times 10^{13}$ Hz
D. $3.0 \times 10^{3}$ Hz
Solution: $E = h\nu = pc$
$\nu = \frac{pc}{h} = \frac{3.3 \times 10^{-29} \times 3 \times 10^8}{6.6 \times 10^{-34}} = 1.5 \times 10^{13}$ Hz
Q12 — Particle Nature of Light–The Photon · medium · numerical
A radio transmitter operates at a frequency 880 kHz and a power of 10 kW. The number of photons emitted per second is
A. $1.72 \times 10^{31}$ ✓ Correct
B. $1.327 \times 10^{25}$
C. $1.327 \times 10^{37}$
D. $1.327 \times 10^{45}$
Solution: $n = \frac{P}{h\nu} = \frac{10^4}{6.63 \times 10^{-34} \times 880 \times 10^3} = 1.72 \times 10^{31}$
Q13 — Particle Nature of Light–The Photon · easy · theory
The energy of a photon of wavelength $\lambda$ is
A. $hc\lambda$
B. $\frac{hc}{\lambda}$ ✓ Correct
C. $\frac{\lambda}{hc}$
D. $\frac{\lambda h}{c}$
Solution: By Planck's quantum theory, the energy of a photon is
$E = h\nu = \frac{hc}{\lambda}$