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#photon emission

8 public questions tagged with this topic.

Photon Energy, Momentum, Power and X-rays

A light source emits \( 1.0 \times 10^{16} \) photons per second with a power of \( 4.0 \, \text{mW} \). What is the wav

**Photon power** light source emits 1.0×10¹⁶ photons/s power 4.0 mW, energy per photon E= P/N =4×10⁻³/10¹⁶=4×10⁻¹⁹ J, wavelength λ= h c/E=6.63×10⁻³⁴×3×10⁸/4×10⁻¹⁹=497 nm, for 6×10¹⁵ photons/s 3.0 mW E=5×10⁻¹⁹ J, for beam 2.5×10¹⁵ photons/s 4×10⁻¹⁹ J each power=10⁻³ W=1 mW, illustrating P = N h f. Energy per photon E = (P/N) = (4.0 × 10⁻³/1.0 × 10¹⁶) = 4.0 × 10⁻¹⁹ J . E = (h c/λ) ⇒ λ = (h c/E) = (6.63 × 10⁻³⁴ × 3 × 10⁸/4.0 × 10⁻¹⁹) ≈ 4.9725 × 10⁻⁷ m = 497.25 nm . Applying E = h f = h c/λ, p = h/λ, K_max = h

Ref: NCERT > Physics Book > Dual Nature of Matter > Photon Energy, Momentum, Power and X-rays

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Atomic Models - Rutherford, Thomson and Bohr

According to Bohr’s model, what happens when an electron in a hydrogen atom transitions from a higher orbit to a lower o

**Thomson's plum pudding model** positive charge uniformly distributed in sphere with electrons embedded, positive charge spread, fails to explain large angle scattering observed. Bohr's model introduces stationary orbits with quantized angular momentum L = n h/2π, physical basis de Broglie standing wave condition 2πr = n λ, circumference fits n wavelengths, explains line spectrum. Bohr’s third postulate states that a photon is emitted with energy equal to the difference between the initial and final energy states. Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and

Ref: NCERT > Physics Book > Atoms and Nuclei > Atomic Models - Rutherford, Thomson and Bohr

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