What EM spectrum region is a 15kW source emitting 2.5×10²² photons/s?

Ultraviolet. Energy per photon=6×10⁻¹⁹J, wavelength=330nm, which lies in the UV range (10–400nm).

How to find wavelength from power and photon rate?

E_photon=P/n=15000/(2.5×10²²)=6×10⁻¹⁹J. Then λ=hc/E=330nm.

15 kW monochromatic source emits 2.5×10²² photons/s — EM spectrum region

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PhysicsModern Physics

QA monochromatic source of light operating at 15 kW emits $2.5\times10^{22}$ photons/s. The region of the EM spectrum to which the emitted radiation belongs is: ($h=6.6\times10^{-34}$ J·s, $c=3\times10^8$ m/s)

A) Microwave B) Infrared C) Visible D) UltravioletModern Physics

MCQ

✅ Correct Answer

D) Ultraviolet

✓

Solution

Find energy per photon

$E_{\text{photon}}=\dfrac{\text{Power}}{\text{photons/s}}=\dfrac{15\times10^3}{2.5\times10^{22}}=6\times10^{-19}\text{ J}$

Find wavelength

$\lambda=\dfrac{hc}{E}=\dfrac{6.6\times10^{-34}\times3\times10^8}{6\times10^{-19}}=\dfrac{1.98\times10^{-25}}{6\times10^{-19}}=3.3\times10^{-7}\text{ m}=330\text{ nm}$

Identify spectrum region

Visible light: 400–700 nm
Ultraviolet (UV): 10–400 nm

$330\text{ nm}<400\text{ nm}$ → Ultraviolet ✓

📘 Photon Energy and EM Spectrum

Photon energy $E=hf=hc/\lambda$. From power and photon rate: $E_{\text{photon}}=P/N$. Then wavelength $\lambda=hc/E$. Compare with spectrum boundaries: UV (10–400 nm), visible (400–700 nm), IR (700 nm–1 mm).

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Important Concepts

Photon Energy from PowerPower = (photons/s) × (energy/photon). So E_photon = P/n = 15kW/(2.5×10²²) = 6×10⁻¹⁹ J.

Wavelength Calculationλ=hc/E = (6.6×10⁻³⁴×3×10⁸)/(6×10⁻¹⁹) = 1.98×10⁻²⁵/6×10⁻¹⁹ = 3.3×10⁻⁷m = 330nm.

EM Spectrum RangesGamma: <0.01nm, X-ray: 0.01–10nm, UV: 10–400nm, Visible: 400–700nm, IR: 700nm–1mm, Microwave: 1mm–1m, Radio: >1m.

Energy-Wavelength RelationshipHigher energy → shorter wavelength. E=6×10⁻¹⁹J ≈ 3.75 eV. UV photons typically have 3.1–124 eV. 3.75 eV is near UV, consistent with 330nm.

FAQs

What is the formula for photon energy?

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E=hf=hc/λ. Here h=6.6×10⁻³⁴ J·s, c=3×10⁸ m/s.

How to convert to eV?

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E=6×10⁻¹⁹ J / 1.6×10⁻¹⁹ J/eV = 3.75 eV. UV photons range from about 3.1 eV (400nm) to 12.4 eV (100nm).

Why not visible?

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Visible range: 400–700nm. Our answer 330nm is below 400nm — it's in the UV range, not visible.

Is UV harmful?

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Yes. UV-C (100–280nm) and UV-B (280–315nm) can cause sunburn and DNA damage. 330nm is UV-A (315–400nm), the least harmful UV range.

Is this from JEE Main 2026?

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Yes, this appeared in JEE Main 2026.

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