Question

Problem 4. Radiation from the black hole accretion disk.] In this problem you will find the temperature of an accretion disk surrounding the black hole, and the wavelength of maximal radiation from the disk. Note that the calculation is super-simplified - if you take more advanced astronomy courses, you will do the same calculation more carefully. In this problem 'BH' means black hole. We will consider the BH of mass equal to twenty solar masses MBH = 20 MSun. (a) Consider a single hydrogen atom at rest infinitely far from the BH. As it falls towards the BH its speed will increase. Find the speed of the atom at distance d = 105 km from the center of the BH. This should be easy enough, as this speed is exactly equal to the escape speed at distance d from the black hole. You already used the equation for escape speed in HW 2 and in Problem 2 of this HW. (b) Find the kinetic energy K = m/v2/2 of the hydrogen atom at that point, with speed v equal to what you just found in part (a). You will have to look up the mass of hydrogen atom my (c) All atoms falling in towards the BH will collide, exchange energy, and form an accretion disk. Since energy is conserved, at distance d the average energy (E) of all particles must remain exactly equal to what you calculated in part (b), i.e. (E) = K. Taking this into account, we write the equation relating the average particle energy in the ideal gas to the gas temperature as 3 K = 7kBT where kp is Boltzmann's constant (look up its numerical value). Using this equation, find the temperature T of the accretion disk at that point. Temperature T you will get from this equation will be in Kelvins. (d) Now that you have found the temperature, use Wien's displacement law Amar = b/T to find the maximum wavelength Amar of thermal radiation of the accretion disk. You already used Wien's law in HW 3 - recall that b is Wien's constant. What is the wavelenght of Imar in nanometers, and in which part of the spectrum is that wavelength?

Answer 1

V= kg = 4x103 kg d=105 km = 10 PM 1 This is in gamma (a) escape speed at dintana d is given by 26 M d we have m=20 Maun = 20x 2 x 1030 m and v= 2X6,67% 10-" 4x1031 108 J53,3641012 = 7.3x109 m/ (6) K = k= J 2K 2x4. 45410lt K-KOT TE 341,38% 10-23 T= 2,15% 109K (d) amat Б 2.898X1023 [; b= 2.6987103 N May2 1 mass of it atom mu- 1.67*10-27 27 kg 2 x 1067410-27 + (7,34 10672 Ik = 4.458 10-14 3kB T 2. 15 x 109 mk] amax = 1,35% 10-12 m part of spectrum

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