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#tube voltage

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Photon Energy, Momentum, Power and X-rays

The maximum frequency of X-rays produced by a \( 20 \, \text{kV} \) tube is: (Take \( h = 6.63 \times 10^{-34} \, \text{

**X-rays from tube** maximum frequency f_max = e V/h, minimum wavelength λ_min = h c/(e V), Duane-Hunt law, for 10 kV f_max=1.6×10⁻¹⁹×10⁴/6.63×10⁻³⁴=2.41×10¹⁸ Hz, λ_min= c/f_max=0.124 nm, for 35 kV λ_min=1240/(35000) eV·nm? Actually 35 keV photon λ=1240/35000=0.0354 nm, for 15 kV 0.0827 nm, for 25 kV 0.0496 nm, for 30 kV 0.0413 nm. Energy E = e V = 1.6 × 10⁻¹⁹ × 20 × 10³ = 3.2 × 10⁻¹⁵ J . vₘₐₓ = (E/h) = (3.2 × 10⁻¹⁵/6.63 × 10⁻³⁴) ≈ 4.83 × 10¹⁸ Hz . Applying E = h f = h c/λ, p = h/λ, K_max = h f - Φ, Φ =

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

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Photon Energy, Momentum, Power and X-rays

The minimum wavelength of X-rays from a \( 15 \, \text{kV} \) tube is: (Take \( h = 6.63 \times 10^{-34} \, \text{J s} \

**X-rays from tube** maximum frequency f_max = e V/h, minimum wavelength λ_min = h c/(e V), Duane-Hunt law, for 10 kV f_max=1.6×10⁻¹⁹×10⁴/6.63×10⁻³⁴=2.41×10¹⁸ Hz, λ_min= c/f_max=0.124 nm, for 35 kV λ_min=1240/(35000) eV·nm? Actually 35 keV photon λ=1240/35000=0.0354 nm, for 15 kV 0.0827 nm, for 25 kV 0.0496 nm, for 30 kV 0.0413 nm. E = e V = 1.6 × 10⁻¹⁹ × 15 × 10³ = 2.4 × 10⁻¹⁵ J . λₘiₙ = (h c/E) = (6.63 × 10⁻³⁴ × 3 × 10⁸/2.4 × 10⁻¹⁵) ≈ 8.2875 × 10⁻¹¹ m = 0.082875 nm . Applying E = h f = h c/λ, p = h/λ, K_max

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

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Photon Energy, Momentum, Power and X-rays

The maximum frequency of X-rays from a \( 35 \, \text{kV} \) tube is: (Take \( h = 6.63 \times 10^{-34} \, \text{J s} \)

**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. E = e V = 1.6 × 10⁻¹⁹ × 35 × 10³ = 5.6 × 10⁻¹⁵ J . vₘₐₓ = (E/h) = (5.6 × 10⁻¹⁵/6.63 × 10⁻³⁴) ≈ 8.446 × 10¹⁸ Hz . Applying E = h f = h c/λ, p = h/λ, K_max = h f - Φ, Φ = h f₀ = h c/λ₀, λ = h/√(2

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

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