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PHYSICS, M7 2023 HSC 23c

The James Webb Space Telescope observed an exoplanet emitting a peak wavelength of 1.14 \(\times\) 10\(^{-5}\) m.

Calculate the temperature of the exoplanet.  (2 marks)

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\(T= 254\) \( \text{K}\)

Show Worked Solution

\(T\) \(=\dfrac{b}{\lambda_{\text{max}}}\)  
  \(=\dfrac{2.898\times 10^{-3}}{1.14 \times 10^{-5}}\)  
  \(= 254\) \( \text{K}\)  

PHYSICS, M7 2023 HSC 9 MC

The graph shows the relationship between radiation intensity and wavelength for a black body at 4500 K.
 

Which statement describes the expected difference in the graph for a black body at 4000 K?

  1. Intensity at all wavelengths will be less.
  2. Intensity at all wavelengths will be greater.
  3. The peak intensity will occur at a higher frequency.
  4. The peak intensity will occur at a shorter wavelength.
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\(A\)

Show Worked Solution

→ The total power output of a black body diminishes with decreasing temperature, resulting in lower intensity across all wavelengths for the 4000 K curve.

\(\Rightarrow A\)

PHYSICS, M7 EQ-Bank 14 MC

The graph shows the electromagnetic radiation emitted from a black body cavity.
 

What is the best estimate of the temperature of this black body?

  1. `5.9 × 10^(3) \ text{K}`
  2. `7.2 × 10^(3) \ text{K}`
  3. `1.7 × 10^(5) \ text{K}`
  4. `5.9 × 10^(6) \ text{K}`
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`A`

Show Worked Solution

→ The peak wavelength of this black body is approximately 490 nm.

`lambda_(max)` `=(b)/(T)`  
`T` `=(2.898 xx10^(-3))/(490 xx10^(-9))`  
  `=5.91 xx10^3  text{K}`   

 
`=>A`

PHYSICS, M7 2018 HSC 2 MC

Which graph is consistent with predictions resulting from Planck's hypothesis regarding radiation from hot objects?
 

 

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`B`

Show Worked Solution

→ Planck’s hypothesis is consistent with Wein’s Law,  `lambda_(max)=(b)/(T)`, or  `lambda_(max) prop (1)/(T)`.

→ So, hotter objects have a greater peak wavelength.

`=>B`

PHYSICS M7 2022 HSC 9 MC

The radiation emitted by a black body has a peak wavelength of  `5.8 xx10^(-7) \ text{m}`.

What is its temperature?

  1. `3000 \ text{K}`
  2. `4500 \ text{K}`
  3. `5000 \ text{K}`
  4. `5500 \ text{K}`
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`C`

Show Worked Solution
`lambda_(max)` `=(b)/(T)`  
`T` `=(b)/(lambda_(max))`  
  `=(2.898 xx10^(-3))/(5.8 xx10^(-7))`  
  `~~5000  text{K}`  

 
`=>C`

PHYSICS, M7 2019 HSC 6 MC

Which graph correctly shows the relationship between the surface temperature of a black body `(T)` and the wavelength `(λ)` at which the maximum intensity of light is emitted?
 

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`A`

Show Worked Solution

Since  `lambda_(max) = (b)/(T), \ \ lambda prop (1)/(T)`  (Wien’s Displacement Law)

`=>A`

PHYSICS, M7 2020 HSC 26

  1. Describe the difference between the spectra of the light produced by a gas discharge tube and by an incandescent lamp.   (2 marks)
  1. The graph shows the curves predicted by two different models, `X` and `Y`, for the electromagnetic radiation emitted by an object at a temperature of 5000 K.
     

    Identify an assumption of EACH model which determines the shape of its curve.   (2 marks)

  1. The diagram shows the radiation curve for a black body radiator at a temperature of 5000 K.
     


     
    On the same diagram, sketch a curve for a black body radiator at a temperature of 4000 K and explain the differences between the curves.   (4 marks)

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a.   The spectra of light produced by a gas discharge tube will consist of lines only at a few discrete wavelengths.

The spectra of light produced by an incandescent lamp will be a continuous spectrum.

b.   Model `X` black bodies absorb and emit energy continuously.

Model `Y` assumes that black bodies absorb and emit energy in discrete quantities.

c.


 
Using `lambda_(max)=(b)/(T)\ \ =>\ \ lambda prop (1)/(T)`

Therefore, the 4000 K curve will have a peak wavelength greater than the 5000 K curve.

The area under the curve and the intensity at all wavelengths will be less for the 4000 K curve, as the total power output of a black body decreases as its temperature decreases.

Show Worked Solution

a.    Differences:

→ The spectra of light produced by a gas discharge tube will consist of lines only at a few discrete wavelengths.
→ The spectra of light produced by an incandescent lamp will be a continuous spectrum.


♦ Mean mark (a) 39%.

b.    Assumptions of EACH model:

→ Model `X` black bodies absorb and emit energy continuously.
→ Model `Y` assumes that black bodies absorb and emit energy in discrete quantities.


♦ Mean mark (b) 44%.

c.

          

Using `lambda_(max)=(b)/(T)\ \ =>\ \ lambda prop (1)/(T)`
Therefore, the 4000 K curve will have a peak wavelength greater than the 5000 K curve.
The area under the curve and the intensity at all wavelengths will be less for the 4000 K curve, as the total power output of a black body decreases as its temperature decreases.

PHYSICS, M7 2020 HSC 22

A capsule travelling at 12 900 m s ¯1 enters Earth's atmosphere, causing it to rapidly slow down to 400 m s ¯1.

  1. During this re-entry, the capsule reaches a temperature of 3200 K.
  2. What is the peak wavelength of the light emitted by the capsule? (2 marks)
  1. Outline TWO limitations of applying special relativity to the analysis of the motion of the capsule.  (3 marks)
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a.   `lambda_(max)=9xx10^(-7)` m

b.   Limitations:

→ The speed of the capsule is not close to the speed of light and so the effects of special relativity are insignificant.

→ The capsule is decelerating and so it is in a non-inertial frame of reference, therefore special relativity is not applicable.

Show Worked Solution
a.    `lambda_(max)` `=(b)/(T)`
    `=(2.898 xx10^(-3))/(3200)`
    `=9.056xx10^(-7)  text{m}`
    `=9xx10^(-7)  text{m}`

 

b.   Limitations:

→ The speed of the capsule is not close to the speed of light and so the effects of special relativity are insignificant.

→ The capsule is decelerating and so it is in a non-inertial frame of reference, therefore special relativity is not applicable.


♦ Mean mark (b) 42%.

PHYSICS, M7 2021 HSC 11 MC

What is the peak wavelength of electromagnetic radiation emitted by a person with a body temperature of 37°C (310 K)?

  1. `9.3 × 10^(-6)\ `m
  2. `7.8 × 10^(-5)\ `m
  3. `9.3 × 10^(-3)\ `m
  4. `7.8 × 10^(-2)\ `m
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`A`

Show Worked Solution
`lambda` `=(b)/(T)`  
  `=(2.898 xx10^(-3))/(310)`  
  `=9.3 xx10^(-6)`  

 
`=>A`