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

By B. Carroll, D. Ostlie

Show description

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

Best introduction books

Āryabhaṭīya of Āryabhaṭa: Critically Edited with Introduction and English Translation

Severe version with English translation of Aryabhatiya, an old Indian textual content in Sanskrit on Astronomy and arithmetic.

Ruled Varieties: An Introduction to Algebraic Differential Geometry

Governed kinds are unions of a kinfolk of linear areas. they're gadgets of algebraic geometry in addition to differential geometry, specifically if the ruling is developable. This ebook is an advent to either facets, the algebraic and differential one. ranging from very user-friendly evidence, the required strategies are built, specifically pertaining to Grassmannians and basic kinds in a model compatible for complicated projective algebraic geometry.

Supportive and Palliative Care in Cancer: an Introduction

This booklet presents a transparent method of developing a consumer involvement approach in a healthcare corporation and its power influence on melanoma prone. utilizing a device package kind strategy drawing on examples of winning prior initiatives and case reports to supply proof of fine perform it describes tips to plan and enforce assorted levels of consumer involvement allowing firms to attract on consumer adventure and services to judge improve and increase the standard of provider that they supply.

Additional info for An Introduction to Modern Astrophysics 2nd ed - SOLUTIONS MANUAL

Sample text

That is, for every neutral hydrogen atom in the n D 3 state, there are 19 in the n D 2 state and 3:55 109 in the n D 1 state. The number of neutral hydrogen atoms per unit area is then N1 hNa i D 5:35 1029 m 2 : N3 The ratio of the number of ionized to neutral atoms is obtained from the Saha equation, Eq. 4). The result shows that NII =NI D 1:33 10 4, so nearly all of the hydrogen atoms are neutral. Thus the above result gives the total number of hydrogen atoms per unit area, N D 5:35 1029 m 2 .

500 nm/ for each wavelength. 594 nm line: Now calculate log10 Œf . 04, so Na D 1:10 1019 m 2 . 22. 1, all of the points lie on the curve of growth. W = /. 9 nm line: Use these with the general curve of growth for the Sun, Fig. 22, to obtain a value of log10 Œf Na . =500 nm/ for each wavelength. 26. log10 Œf hNa i. W = / 4:58 4:69 3:90 4:02 Now calculate log10 Œf . 9 nm line: The average value of log10 Na for this problem is hNa i D 20:18, so hNa i D 1:51 1020 m 2 . This is the value of the number of neutral hydrogen atoms per unit area with an electron in the n D 3 orbit.

14), E1 D 13:6 eV, E2 D 3:40 eV, and E3 D 1:51 eV, and gn D 2n2 for hydrogen. 5, we find N3 =N2 D 5:14 10 2 and N3 =N1 D 2:82 10 10. That is, for every neutral hydrogen atom in the n D 3 state, there are 19 in the n D 2 state and 3:55 109 in the n D 1 state. The number of neutral hydrogen atoms per unit area is then N1 hNa i D 5:35 1029 m 2 : N3 The ratio of the number of ionized to neutral atoms is obtained from the Saha equation, Eq. 4). The result shows that NII =NI D 1:33 10 4, so nearly all of the hydrogen atoms are neutral.

Download PDF sample

Rated 4.01 of 5 – based on 30 votes