Q27. A metal surface emits 4.16 × 1016 photons in 1 second. This monochromatic light emits photoelectrons (in A) in the second. When this light incidents on metal strip, (round off to one decimal place) of the emitted photoelectrons.
Given: Work function of the metal = 2.0 eV
Charge of an electron = 1.6 × 10-19 C
Planck’s constant = 6.626 × 10-34 Js
The photoelectric effect explains how light ejects electrons from metals, crucial for CSIR NET Life Sciences and physics exams. When 16 mW monochromatic light emits 1015 photons per second on a metal strip, calculating the wavelength reveals photon energy for photoelectron emission.
Problem Solution
The power of the light source equals the total energy delivered by all photons per second. With power P = 16 mW = 1.6 × 10-2 W and n = 1015 photons per second, the energy per photon is E = P/n = 1.6 × 10-17 J.
The photon energy relates to wavelength via E = hc/λ, where h = 6.626 × 10-34 J s and c = 3 × 108 m/s, so λ = hc/E = 1.242 × 10-8 m = 124.2 Å (after converting to angstroms and rounding to one decimal place).
This wavelength determines if photoelectrons emit: if λ falls below the metal’s threshold wavelength, photons carry enough energy to overcome the work function, per Einstein’s photoelectric equation hν = ϕ + Kmax, where charge e = 1.6 × 10-19 C and electron mass m = 9.1 × 10-31 kg inform related calculations like stopping potential or velocity.
Step-by-Step Derivation
- Convert power: 16 mW = 0.016 W.
- Energy per photon: E = 0.016 / 1015 = 1.6 × 10-17 J.
- Wavelength: λ = (6.626 × 10-34 × 3 × 108) / 1.6 × 10-17 = 1.242 × 10-8 m.
- In angstroms: 1.242 × 10-8 × 1010 = 124.2 Å.
No work function is provided, so the query focuses solely on incident light wavelength, not emission kinetics.
Key Formula Applications
Photon energy E = hc/λ, where power P = nE yields λ = nhc/P. Here, P = 1.6 × 10-2 W, n = 1015 s-1, giving E = 1.6 × 10-17 J and λ = 124.2 Å.
Electron charge 1.6 × 10-19 C and mass 9.1 × 10-31 kg support follow-up: maximum velocity v = √(2Kmax/m), where Kmax = E – ϕ.
Exam Relevance for CSIR NET
CSIR NET questions test this: power-to-wavelength conversion verifies threshold conditions. Similar problems use saturation current I = ne for photoelectrons per second.
Quick Summary
- Final Answer: Wavelength = 124.2 Å
- Key Constants: h = 6.626 × 10-34 J s, e = 1.6 × 10-19 C
- Application: Photoelectric threshold determination
