2022 A/L Physics Past Paper · Question 08 (English Medium)

Answer: (2) $300\ \text{nm}$

What the question asks

We want: the peak wavelength of a blackbody at $10\,000\ \text{K}$, when the sun at $6000\ \text{K}$ peaks at $500\ \text{nm}$.

What each letter means:

$\lambda_m$ ($\text{nm}$): wavelength of the peak of the radiation.

$T$ ($\text{K}$): absolute temperature of the blackbody.

$C$ ($\text{m K}$): the constant in Wien's displacement law.

Relevant theory

In the spectral distribution of black body radiation, the product of the wavelength

relevant to the maximum intensity and the relevant absolute temperature is constant.

Required equations

$\lambda_m T = C$

Reasoning and calculation

What stays the same: the constant $C$, for every blackbody.

What changes: the temperature $T$, and with it the peak wavelength $\lambda_m$.

For example, watch an electric stove ring as it warms up.

At first it glows a dull red. As it gets hotter it turns orange, then yellowish white.

Red light has a long wavelength. Yellow-white light has a shorter wavelength.

So a hotter body peaks at a shorter wavelength.

$$\lambda_m T = C$$

In words: peak wavelength times absolute temperature is always the same constant.

If $T$ goes up, $\lambda_m$ must come down by the same factor.

The constant $C$ is the same for the sun and for the second blackbody.

$$\lambda_1 T_1 = \lambda_2 T_2$$

Solve for the new peak wavelength:

$$\lambda_2 = \dfrac{\lambda_1 T_1}{T_2}$$

Put in the numbers:

$$\lambda_2 = \dfrac{500\ \text{nm} \times 6000\ \text{K}}{10\,000\ \text{K}}$$

$$\lambda_2 = 300\ \text{nm}$$

Reading the options

The value $300\ \text{nm}$ is option (2).

Option (5), $800\ \text{nm}$, comes from using the ratio upside down.

Checks

Dimension check:

The unit of $\lambda_2$ is $\text{nm} \times \text{K} \div \text{K} = \text{nm}$, a length, as it should be.

$(l1 * t1) / t2$

Does it make sense?

The temperature went up from $6000\ \text{K}$ to $10\,000\ \text{K}$.

So the peak wavelength must go down. $300\ \text{nm}$ is less than $500\ \text{nm}$.

Conclusion

$\lambda_2 = 300\ \text{nm}$

$\lambda_2 = \dfrac{\lambda_1 T_1}{T_2} = \dfrac{500 \times 6000}{10\,000} = 300\ \text{nm}$.

The answer is (2).

Hotter means a shorter peak wavelength. The product $\lambda_m T$ never changes.

Write $\lambda_1 T_1 = \lambda_2 T_2$ and solve.

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