Skip to content
Login Register

#photon frequency

8 public questions tagged with this topic.

Photon Energy, Momentum, Power and X-rays

The threshold frequency of a metal is \( 4.8 \times 10^{14} \, \text{Hz} \). What is the maximum kinetic energy for ligh

**Photon energy** E = h f = h c/λ, momentum p = h/λ = E/c, power P = N E/t, N number of photons per second, wavelength λ = h c/E, for photon momentum 1.5×10⁻²⁷ kg·m/s frequency f = p c/h? Actually E = p c, f = E/h = p c/h =1.5×10⁻²⁷×3×10⁸/6.63×10⁻³⁴=6.79×10¹⁴ Hz, energy E= h f =4.5×10⁻¹⁹ J, for 5.0×10⁻¹⁹ J frequency f=E/h=7.54×10¹⁴ Hz. Φ₀ = h v₀ = 6.63 × 10⁻³⁴ × 4.8 × 10¹⁴ = 3.1824 × 10⁻¹⁹ J . E = h v = 6.63 × 10⁻³⁴ × 6.8 × 10¹⁴ = 4.5084 × 10⁻¹⁹ J . Kₘₐₓ = E

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

View Details
Photoelectric Effect - Work Function and Threshold

Which property of incident light determines the maximum kinetic energy of photoelectrons, assuming the metal remains the

**Photoelectric effect** demonstrates particle nature of light, photon energy E = h f, work function Φ = h f₀ minimum energy to eject electron, threshold frequency f₀ = Φ/h, threshold wavelength λ₀ = hc/Φ, hc=1240 eV·nm, maximum kinetic energy K_max = h f - Φ = e V₀, V₀ stopping potential, photocurrent proportional to intensity (number of photons) when f>f₀, saturation current depends on intensity, frequency determines K_max not current. According to Einstein’s photoelectric equation, Kₘₐₓ = h v - Φ₀ , the maximum kinetic energy depends on the frequency of the incident light. Applying E = h f

Ref: NCERT > Physics Book > Dual Nature of Matter > Photoelectric Effect - Work Function and Threshold

View Details
Bohr Model Energy Levels and Hydrogen Spectrum

What is the frequency of a photon emitted when an electron drops from \( n = 3 \) to \( n = 1 \) in a hydrogen atom? (Us

**Excitation** energy required to go from n=1 to n=3 is 12.09 eV, from ground to n=∞ ionization 13.6 eV, state n=∞ means ionized, electron free with zero energy, highest level reached by electron beam energy determines which levels can be excited, e.g., 11 eV beam from ground can reach n=2 (10.2 eV) but not n=3 (12.09 eV), so max n=2. Δ E = 12.09 eV = 1.9344 × 10⁻¹⁸ J . nu = (Δ E/h) = (1.9344 × 10⁻¹⁸/6.6 × 10⁻³⁴) ≈ 2.93 × 10¹⁵ Hz . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE

Ref: NCERT > Physics Book > Atoms and Nuclei > Bohr Model Energy Levels and Hydrogen Spectrum

View Details
Bohr Model Energy Levels and Hydrogen Spectrum

What is the frequency of a photon emitted when an electron drops from \( n = 2 \) to \( n = 1 \) in a hydrogen atom? (Us

**Hydrogen transitions** example n=3→n=1 ΔE=13.6(1-1/9)=12.09 eV, photon 12.09 eV, λ=1240/12.09≈102.6 nm Lyman series, n=3→n=2 ΔE=1.89 eV Balmer visible Hα 656 nm. Absorption photon energy must match difference, if atom in n=2 absorbs 1.89 eV jumps to n=3, if absorbs 12.75 eV from ground 1→4 because -13.6+12.75=-0.85 eV = -13.6/16. Δ E = 10.2 eV = 1.632 × 10⁻¹⁸ J . nu = (Δ E/h) = (1.632 × 10⁻¹⁸/6.6 × 10⁻³⁴) ≈ 2.47 × 10¹⁵ Hz . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1 u = 931.5 MeV, evaluation yields

Ref: NCERT > Physics Book > Atoms and Nuclei > Bohr Model Energy Levels and Hydrogen Spectrum

View Details