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SOLUTIONS TO EXERCISE 20
(a) The redshift of the quasar LPIF13 is z = 2.58 and we know that delta(lambda) z = ------------- . For (lambda) = 656.3 nm, the hydrogen-alpha line, (lambda) the line appears at (lambda) + delta(lambda) = 656.3 + (2.58 * 656.3) = 2350 nm. For (lambda) = 486.1 nm, the hydrogen-beta line, the line appears at 486.1 + (2.58 * 486.1) = 1740 nm. (b) For the velocity at which LPIF13 is moving away from Earth, we must use the relativistic Doppler formula (the non-relativistic Doppler formula would give an apparent velocity greater that the speed of light and is thus wrong). 1 + v/c z = sqrt(-------) - 1. Since z = 2.58, we can move the 1 over to get 1 - v/c 1 + v/c (2.58 + 1)^2 = ------- which leads to (12.8) * (1 - v/c) = 1 + v/c 1 - v/c and thus 12.8 - 12.8(v/c) = 1 + (v/c), or 11.8 = 13.8(v/c). So v = c * (11.8)/(13.8) = 2.57 * 10^8 m/sec. So the expansion of the universe is carrying LPIF13 away at about 86% the speed of light. c) When high-density, low-energy clouds of gas are illuminated, they produce absorption lines at the rest wavelengths of the atoms in the clouds. Since this cloud contains hydrogen gas, we know that absorption lines would be produced at rest wavelengths of 656.3 nm (H-alpha) and 486.1 nm (H-beta). Since these clouds are not at rest, but rather at redshift z = 1.5, these absorption lines would be seen at wavelengths 656.3 * (1 + 1.5) = 1641 nm for H-alpha, and 486.1 * (1 + 1.5) = 1215 nm for H-beta. |