Find the temperature of a blackbody if its spectrum has a peak at (a) λmax = 700 nm (visible), (b) λmax = 3 cm

Questions 3-25 · Q25

Q. Find the temperature of a blackbody if its spectrum has a peak at

(a)

\lambda_{max} = 700

nm (visible),

(b)

\lambda_{max} = 3

cm (microwave region) and

(c)

\lambda_{max} = 3

m (short radio waves). (Take Wien's constant

b = 2.897\times10^{-3}

m K)

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Concept understanding — Wien's Displacement Law

Wien's Displacement Law: From Intuition to Formula

You already know that hot objects glow. A piece of iron heated in a forge first turns dull red, then bright orange, then yellow-white, and finally blue-white as it gets hotter. That colour change is not random — it follows a strict physical rule.

The Core Intuition

Think of temperature as a measure of how violently the atoms in an object are jiggling. When atoms vibrate faster (higher temperature), they emit electromagnetic waves with shorter wavelengths. Shorter wavelength means higher frequency, which means more energy per photon.

So as temperature rises, the "peak" of the emitted radiation shifts toward the blue end of the spectrum. As temperature falls, the peak shifts toward the red end. This is why a cool star looks red and a very hot star looks blue-white.

Note

The "peak" here means the wavelength at which the object emits the most intense radiation — the brightest colour in its thermal glow.

The Precise Statement

Wien's Displacement Law gives you the exact relationship between the temperature of a blackbody (an ideal thermal emitter) and the wavelength at which its emission is strongest:

\lambda_{\text{max}} \, T = b

where $b$ is Wien's displacement constant:

b = 2.898 \times 10^{-3} \ \text{m·K}

In most exam problems, you use:

\lambda_{\text{max}} \, T = 2.9 \times 10^{-3} \ \text{m·K}

\lambda_{\text{max}} = \frac{2.9 \times 10^{-3}}{T}

What This Tells You

If you know the temperature of a star, a filament, or any hot object, you can immediately find the wavelength where it shines brightest. Conversely, if you measure the peak wavelength of a star's light, you can calculate its surface temperature.

Example: The Sun's surface temperature is about 5800 K. Its peak wavelength is:

\lambda_{\text{max}} = \frac{2.9 \times 10^{-3}}{5800} \approx 5.0 \times 10^{-7} \ \text{m} = 500 \ \text{nm}

That is green light — right in the middle of the visible spectrum. The Sun actually peaks in green, but our eyes perceive the combined light as white. …

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By Wien's displacement law, T = b/λmax, with b = 2.897×10⁻³ m K.

λmax = 700 nm = 7×10⁻⁷ m: T = 2.897×10⁻³ / 7×10⁻⁷ ≈ 4138.6 K ≈ 4138 K -- a temperature typical of an incandescent lamp filament or a Sun-like star, consistent with the wavelength being in the visible range.
λmax = 3 cm = 0.03 m: T = 2.897×10⁻³ / 0.03 ≈ 0.09657 K -- an extremely low temperature, since microwave-peaked blackbody radiation (like the cosmic microwave background) corresponds to objects only a fraction of a degree above absolute zero. …

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Q.Find the temperature of a blackbody if its spectrum has a peak at

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