By B. Carroll, D. Ostlie

Show description

Read Online or Download An Introduction to Modern Astrophysics 2nd ed - SOLUTIONS MANUAL PDF

Similar introduction books

G. Gordon Gibson, Paul Skett's Introduction to Drug Metabolism (3rd Edition) PDF

The services of the authors of this name is complementary, with one in accordance with biochemistry/toxicology and the opposite in line with pharmacology/medicine. the topic is approached from either biochemical and physiological angles. it really is directed at complex undergraduate biochemists, pharmacologists, pre-clinical clinical scholars and complex undergraduate/postgraduate toxicologists.

Ajanta : monochrome reproductions of the Ajanta frescoes - download pdf or read online

Ajanta; the color and monochrome reproductions of the Ajanta Frescoes in response to images, with an explanatory textual content by way of G. Yazdani, and an Appendix and Inscrition. by means of N. P. Chakravarti. released less than the particular authority of His Exalted Highness the Nizam.

Extra resources for An Introduction to Modern Astrophysics 2nd ed - SOLUTIONS MANUAL

Sample text

A C b v / sec  e D a sec  Z 1 e v sec  v sec  d subscripts for convenience) v Z d 0 v C b sec  1 0 v e v sec  d v: The first integral on the right-hand side is 1= sec Â, while the second can be integrated by parts, resulting in ˇ1 ! 24 Assuming that there is zero flux received from the center of the line (as in the Schuster–Schwarzschild model), the equivalent width is equal to the area of the half-ellipse divided by the semimajor axis. From Eq. 25 Heisenberg’s uncertainty principle, Eq. 20), relates E, the uncertainty in the energy of an atomic orbital, to t, the time an electron occupies the orbital before making a downward transition: E t D h 2 t : When an electron makes a downward transition from an initial to a final orbital, the energy of the emitted photon is (Eq.

2/2 D 8, E1 D 13:6 eV, and E2 D 3:40 eV. Then, with N2 =N1 D 0:01, T D 1:97 104 K. If the number of atoms in the first excited state is only 10% of the number of atoms in the ground state, then T D 3:21 104 K. 6 (a) We solve the Boltzmann equation, Eq. 3/3 D 18 and E3 D N3 =N1 D 1, then T D 6:38 104 K. (b) From the Boltzmann equation using T D 85,400 K, N3 =N1 D N3 =N D 1:74. 42 1:51 eV, we find that if Solutions for An Introduction to Modern Astrophysics (c) 43 As T ! 1, the exponential in the Boltzmann equation goes to unity and Nb =Na !

4 The most probable speed, vmp , occurs at the peak of the Maxwell–Boltzmann distribution, Eq. 1). Setting d nv =dv D 0, we find Á d nv m Á3=2 d 2 e mv =2kT v 2 D 0: D4 n dv 2 kT dv This leads to  à mv 3 2 C 2v e mv =2kT D 0; kT p so that vmp D 2kT =m, which is Eq. 2). 5 Solving the Boltzmann equation, Eq. 2/2 D 8, E1 D 13:6 eV, and E2 D 3:40 eV. Then, with N2 =N1 D 0:01, T D 1:97 104 K. If the number of atoms in the first excited state is only 10% of the number of atoms in the ground state, then T D 3:21 104 K.

Download PDF sample

An Introduction to Modern Astrophysics 2nd ed - SOLUTIONS MANUAL by B. Carroll, D. Ostlie


by Kevin
4.4

Rated 4.16 of 5 – based on 38 votes