Answer:
<em><u>solution</u></em>
<em>3</em><em>0</em><em>8</em><em>=</em><em>2</em><em>0</em><em> </em><em>swings</em><em> </em>
<em> </em><em>?</em><em>:</em><em>:</em><em>:</em><em> </em><em>=</em><em>1</em>
<em>(</em><em> </em><em>3</em><em>0</em><em>8</em><em>×</em><em>1</em><em>)</em><em>÷</em><em>2</em><em>0</em>
<em>3</em><em>0</em><em>8</em><em>÷</em><em>2</em><em>0</em>
<em>1</em><em>5</em><em>4</em><em>÷</em><em>1</em><em>0</em>
<em>=</em><em>1</em><em>5</em><em>.</em><em>4</em>
<em>=</em>15.4
147.09975 newton meters per second
Answer:
b) Betelgeuse would be
times brighter than Sirius
c) Since Betelgeuse brightness from Earth compared to the Sun is
the statement saying that it would be like a second Sun is incorrect
Explanation:
The start brightness is related to it luminosity thought the following equation:
(1)
where
is the brightness,
is the star luminosity and
, the distance from the star to the point where the brightness is calculated (measured). Thus:
b)
and
where
is the Sun luminosity (
) but we don't need to know this value for solving the problem.
is light years.
Finding the ratio between the two brightness we get:
![\displaystyle{\frac{B_{Betelgeuse}}{B_{Sirius}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (26\ ly)^2}{26L_{Sun}} \approx 1.43 \cdot 10^{6} }](https://tex.z-dn.net/?f=%5Cdisplaystyle%7B%5Cfrac%7BB_%7BBetelgeuse%7D%7D%7BB_%7BSirius%7D%7D%3D%5Cfrac%7B10%5E%7B10%7DL_%7BSun%7D%7D%7B4%5Cpi%20%28427%5C%20ly%29%5E2%7D%20%5Ctimes%20%5Cfrac%7B4%5Cpi%20%2826%5C%20ly%29%5E2%7D%7B26L_%7BSun%7D%7D%20%5Capprox%201.43%20%5Ccdot%2010%5E%7B6%7D%20%7D)
c) we can do the same as in b) but we need to know the distance from the Sun to the Earth, which is
. Then
![\displaystyle{\frac{B_{Betelgeuse}}{B_{Sun}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (1.581\cdot 10^{-5}\ ly)^2}{1\ L_{Sun}} \approx 1.37 \cdot 10^{-5} }](https://tex.z-dn.net/?f=%5Cdisplaystyle%7B%5Cfrac%7BB_%7BBetelgeuse%7D%7D%7BB_%7BSun%7D%7D%3D%5Cfrac%7B10%5E%7B10%7DL_%7BSun%7D%7D%7B4%5Cpi%20%28427%5C%20ly%29%5E2%7D%20%5Ctimes%20%5Cfrac%7B4%5Cpi%20%281.581%5Ccdot%2010%5E%7B-5%7D%5C%20ly%29%5E2%7D%7B1%5C%20L_%7BSun%7D%7D%20%5Capprox%201.37%20%5Ccdot%2010%5E%7B-5%7D%20%7D)
Notice that since the star luminosities are given with respect to the Sun luminosity we don't need to use any value a simple states the Sun luminosity as the unit, i.e 1. From this result, it is clear that when Betelgeuse explodes it won't be like having a second Sun, it brightness will be 5 orders of magnitude smaller that our Sun brightness.
1) n=1 -> n=2 : delta E = -5+11 = 6. The answer is D
<span>2) n=1 -> n=3 : delta E = -2+11 = 9. The answer is B </span>
<span>3) n=1 -> n=4 : delta E = -1+11 = 10. No solution available </span>
<span>4) n=1 -> infinity delta E = 11. The answer is A </span>
<span>5) not absorbed would be C, as there is no transition with delta E of 8. </span>