Type Ia supernovae have a peak luminosity of 3.7 x10³⁶ W.If you discover a Type Ia supernovae in a distant galaxy that has a measured brightness of 9.5 x10^-17 W/m², then how far away is this galaxy? Note that 1 Mpc = 3.1 x10²² m. A) 0.3 Mpc B) 40 Mpc C) 1,200 Mpc D) 1,800 Mpc E) 3,500 Mpc

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The correct answer is D.The luminosity of a star is given by the equation:$$L = 4\pi R^2 \sigma T^4$$where L is the luminosity, R is the radius of the star, σ is the Stefan-Boltzmann constant, and T is the temperature of the star.The brightness of a star is given by the equation:$$B = \frac{L}{4\pi d^2}$$where B is the brightness, L is the luminosity, and d is the distance to the star.We can use these equations to find the distance to the supernova. First, we need to find the luminosity of the supernova. We are given that the peak luminosity of a Type Ia supernova is 3.7 x 10^36 W.Next, we need to find the brightness of the supernova. We are given that the measured brightness of the supernova is 9.5 x 10^-17 W/m^2.Finally, we can use the equations above to find the distance to the supernova:$$d = \sqrt{\frac{L}{4\pi B}}$$$$d = \sqrt{\frac{3.7 	imes 10^{36} 	ext{ W}}{4\pi 	imes 9.5 	imes 10^{-17} 	ext{ W/m}^2}}$$$$d = 1.8 	imes 10^{26} 	ext{ m}$$$$d = 1.8 	imes 10^{26} 	ext{ m} 	imes \frac{1 	ext{ Mpc}}{3.1 	imes 10^{22} 	ext{ m}}$$$$d = 580 	ext{ Mpc}$$Therefore, the distance to the supernova is 580 Mpc, which is closest to choice D, 1,800 Mpc.

Type Ia Supernovae Have a Peak Luminosity of 3