alleryStart learning →Photon energy:
- Planck constant alone
- hν (= hc/λ)
- kT
- mc²
Show answer and solution
Correct answer: B. hν (= hc/λ)
We have E = hν.
So E = hc/λ.
h = 6.626 × 10⁻³⁴ J s (Planck's constant).
Higher frequency (shorter wavelength) = higher energy photons.
Indian Navy SSR Dual Nature of Radiation and Matter — practice questions
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Worked practice questions
Threshold wavelength for a metal with work function 3 eV:
- About 413 nm (= 1240/3)
- 100 nm
- 1240 nm
- 700 nm
Show answer and solution
Correct answer: A. About 413 nm (= 1240/3)
We have λ₀ = hc/φ.
So λ₀ = 1240/3 ≈ 413 nm.
Below this wavelength (= higher frequency) photoemission occurs.
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