Question bank
Practice A-Level CAIE Physics 25 2 Stellar Radii questions by syllabus topic with past-paper context, marks, difficulty and question previews on Eduninja.
Question 3
Question 3(c)
The radiant flux intensity of the radiation from the star in (b) is \(2.52 \times 10^{-8} \mathrm{Wm}^{-2}\) when observed at a distance of \(4.16 \times 10^{16} \mathrm{~m}\) from the star.
Question 3(c)(ii)
Determine the radius of the star. radius = m
\[ \begin{aligned} L=4 \pi \sigma r^{2} T^{4} 5.48 \times 10^{26}=4 \pi \times 5.67 \times 10^{-8} \times r^{2} \times 6980^{4} \end{aligned} \] C1 \(r=5.69 \times 10^{8} \mathrm{~m}\) A1
Question 9
Question 9(c)
The star in (b) has a radius of \(2.3 \times 10^{9} \mathrm{~m}\) and a luminosity of \(1.4 \times 10^{28} \mathrm{~W}\). All the energy released from the formation of \({ }_{2}^{4} \mathrm{He}\) is radiated away from the star. All the energy that is radiated from the star has been released in the formation of \({ }_{2}^{4} \mathrm{He}\). Determine:
Question 9(c)(ii)
the surface temperature of the star. temperature = K
\[ L=4 \pi \sigma r^{2} T^{4} \] \[ 1.4 \times 10^{28}=4 \pi \times 5.67 \times 10^{-8} \times\left(2.3 \times 10^{9}\right)^{2} \times T^{4} \] C1 \(T=7800 \mathrm{~K}\) A1
Question 9
Question 9(b)
A star in the constellation Canis Major is a distance of \(8.14 \times 10^{16} \mathrm{~m}\) from the Earth and has a luminosity of \(9.86 \times 10^{27} \mathrm{~W}\). The surface temperature of the star is 9830 K .
Question 9(b)(ii)
Determine the radius of the star. radius = m
\[ \begin{aligned} L=4 \pi \sigma r^{2} T^{4} 9.86 \times 10^{27}=4 \times \pi \times 5.67 \times 10^{-8} \times r^{2} \times 9830^{4} \end{aligned} \] C1 radius \(=1.22 \times 10^{9} \mathrm{~m}\) A1
Question 9(c)
Explain how the surface temperature of a distant star may be determined from the wavelength spectrum of the light from the star.
wavelength of peak intensity determined (from spectrum of star) B1 wavelength of peak intensity from object of known temperature determined B1 Wien's displacement law used or wavelength of peak intensity inversely proportional to temperature B1
Question 10
Question 10(a)
The Sun has a surface temperature of 5780 K . The luminosity of the Sun is \(3.85 \times 10^{26} \mathrm{~W}\).
Question 10(a)(i)
Calculate the radius of the Sun. radius = m
\[ L=4 \pi \sigma r^{2} T^{4} \] \[ 3.85 \times 10^{26}=4 \pi \times 5.67 \times 10^{-8} \times r^{2} \times 5780^{4} \] C1 \(r=6.96 \times 10^{8} \mathrm{~m}\) A1
Question 10(a)(iii)
The variation with wavelength of the intensity of radiation emitted from the Sun is shown in Fig. 10.1. Another star has the same radius as the Sun but has a lower surface temperature. On Fig. 10.1, sketch a line to show the variation with wavelength of the intensity of the radiation emitted for this star.
line of same shape showing peak intensity at greater wavelength B1 line of same shape showing lower peak intensity B1
Question 10
Question 10(a)
State Wien's displacement law.
wavelength of maximum intensity is inversely proportional to (thermodynamic) temperature B1
Question 10(b)
Fig. 10.1 shows the wavelength distributions of electromagnetic radiation emitted by two stars A and B . The surface temperature of star A is known to be 5800 K .
Question 10(b)(i)
Determine the surface temperature of star B . surface temperature = K
\(\lambda_{\text {MAX }}=0.50 \mu \mathrm{~m}\) for A and \(0.65 \mu \mathrm{~m}\) for B C1 \[ \begin{aligned} T =5800 \times(0.50 / 0.65) =4500 \mathrm{~K} \end{aligned} \] A1
Question 10(b)(ii)
Star B appears less bright than star A when viewed from the Earth. Use Fig. 10.1 to suggest, with a reason, how else the physical appearance of star B compares with that of star A.
(star B has) greater peak / average wavelength B1 (star B looks) redder B1
Question 10
Question 10(b)
The Sun has a radius of \(6.96 \times 10^{8} \mathrm{~m}\) and a surface temperature of 5780 K . Light from the Sun is observed to have a peak intensity at a wavelength of 501 nm .
Question 10(b)(ii)
Another star emits radiation that has a peak intensity at a wavelength of 624 nm . Determine the surface temperature of this star. surface temperature =
\(\lambda_{\text {max }} T=\) constant C1 temperature \(=(5780 \times 501) / 624\) \[ \text { = } 4640 \mathrm{~K} \] A1
Question 12
Question 12(b)
The Sun has a surface temperature of 5800 K . The wavelength \(\lambda_{\max }\) of light for which the maximum rate of emission occurs from the Sun is 500 nm . The scientist observing the star in (a) finds that the wavelength for which the maximum rate of emission occurs from the star is 430 nm .
Question 12(b)(i)
Show that the surface temperature of the star in (a) is approximately 6700 K . Explain your reasoning.
(Wien's displacement law states) \(\lambda_{\text {max }} \propto 1 / T\) B1 so \(T=(5800 \times 500) / 430=6700 \mathrm{~K}(6740 \mathrm{~K})\) A1
Question 12(b)(ii)
Use the information in (a) and (b)(i) to determine the radius of the star. radius = m
\[ \begin{aligned} L=4 \pi \sigma r^{2} T^{4} 4.8 \times 10^{29}=4 \pi \times 5.67 \times 10^{-8} \times 6700^{4} \times r^{2} \end{aligned} \] C1 radius \(=1.8 \times 10^{10} \mathrm{~m}\) A1
Question 10
Question 10(a)
State Wien's displacement law. Identify any symbols that you use.
temperature inversely proportional to wavelength M1 temperature is thermodynamic temperature of surface, and wavelength is the wavelength at which maximum emission rate occurs A1
Question 12
Question 12(d)
The radius of the Sun is \(6.96 \times 10^{8} \mathrm{~m}\). Show that the temperature T of the surface of the Sun is 5770 K .
\[ \begin{aligned} \mathrm{L}=4 \pi \sigma r^{2} \mathrm{~T}^{4} 3.83 \times 10^{26}=4 \times \pi \times 5.67 \times 10^{-8} \times 6.96 \times 10^{82} \times \mathrm{T}^{4} \text { leading to } \mathrm{T}=5770 \mathrm{~K} \end{aligned} \] B1
Question 12(e)
The wavelength \(\lambda_{\text {max }}\) of light for which the maximum rate of emission occurs from the Sun is \(5.00 \times 10^{-7} \mathrm{~m}\). The temperature of the surface of the star Sirius is 9940 K . Use information from (d) to determine the wavelength of light for which the maximum rate of emission occurs from Sirius. wavelength = m
\[ \begin{aligned} \lambda_{(\max )} \propto \frac{1}{\mathrm{~T}} \frac{5.00 \times 10^{-7}}{\lambda}=\frac{9940}{5770} \end{aligned} \] C1 \(\lambda=2.90 \times 10^{-7} \mathrm{~m}\) A1