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Cosmology (Physics 20B) Homework 3
Professor Cooper (FRH 2123)
Name:
Student ID #:
Discussion Section:
Always show your work and write legibly. Designate your final answers (when numeric) by circling/outlining
them. Staple all pages together including this coverpage; it’s not recommended that you attempt to fit your
entire answer in the space available on this coverpage. Finally, all times (years, minutes, seconds, etc.) are
given in Earth units, unless otherwise stated.
Completed assignments may be turned in via the P20B drop-box located in the hallway area outside of FRH
4129.
**As with all course material for Physics 20B (including online lecture notes), this homework is not to be
reproduced, redistributed, or sold in any form.**
1. Using Einsteins famous E = mc2 equation, show that when 4 protons fuse into 1 helium nucleus within
the Sun roughly 4.14 × 10−12 Joules of energy is released. (The mass of one individual proton is 1.6725 ×
10−27 kg; the mass of one helium nucleus is 6.644 × 10−27 kg.)
The amount of energy in one proton before the reaction is given by:
Eproton = mproton c2 .
Because there are 4 protons before the reaction, the total energy stored by protons is 4 times this value:
Eall protons = 4mproton c2 .
The amount of energy in the helium nucleus after the reaction is given by:
EHelium = mHelium c2 .
The amount of energy released is the difference between the energy in the Helium nucleus and the energy
in all the protons:
Ereleased = 4mproton c2 − mHelium c2 .
Plugging in the given masses and the speed of light (c = 3 × 108 m/s) gives us the final answer:
Ereleased = 4×1.6725 ×10−27 kg×(3×108 m/s)2 −6.644×10−27 kg×(3×108 m/s)2 = 4.14×10−12 J.
The amount of energy released in one fusion reaction is 4.14 × 10−12 J.
2. The total luminosity (energy radiated per second) of the Sun is about 4 × 1026 J/s. Given the answer from
the previous question, how many nuclei of helium are being created per second in the Sun?
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The total amount of energy released per second in the sun (in Joules per second) is the amount of energy
released per reaction (in Joules) times the number of reactions per second:
Energy
Energy
reactions
=
×
.
second
reaction
second
Because each reaction gives us one Helium nucleus, the number of Helium nuclei produced per second
is the number of reactions per second, so we simply need to solve the above equation for the number of
reactions per second to get the number of Helium nuclei created per second.
Energy
second
Helium nuclei
;
=
Energy
second
reaction
Helium nuclei
4 × 1026 J/s
= 9.66 × 1037 .
=
second
4.14 × 10−12 J
The number of Helium nuclei being produced per second is 9.66 × 1037 .
3. You are 1 meter away from a candle and 10 meters away from a 500 Watt light bulb, but both the candle
and the bulb look to be the same brightness to you. How luminous is the candle (in Watts)?
The apparent brightness of the light bulb can be determined from the brightness formula:
b=
L
.
4πR2
Here b is the apparent brightness of the bulb, L is the luminosity of the bulb (500 W), and R is your distance from the bulb (10 m). Plugging in these numbers tells us that the bulb’s brightness is 0.398 W/m2 .
Becuase we know that the bulb and the candle appear as the same brightness to you, this is the brightness
of the candle also. Based on this brightness and your distance from the candle, we can find the candle’s
luminosity:
2
× b;
Lcandle = 4πRcandle
2
2
Lcandle = 4π(1 m
m
.
) × 0.398 W/
The luminosity of the candle is 5 W.
4. An alien spaceship is racing towards the Earth at a speed of 60, 000 km/s and shines a laser beam at the
Earth. On Earth, we observe the light in the laser beam to be green (i.e. λobserved = 5500 Å). What is the
“rest wavelength” of the laser?
Because we have light coming from a moving ship, we know that there is going to be a Doppler shift.
The Doppler shift equation tells us how the wavelength is going to change because of this movement1 :
∆λ
v
= .
λ0
c
1
Some of you have observed a problem here: this equation only applies when v is much less than the speed of light, whereas
v is 20% of the speed of light in this problem. However, because we haven’t given you the more general formula, you can assume
that the speed of the ship is much less than the speed of light in this case. If the more precise formula is applied the answer is only
3% different.
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In this equation, λ0 is the rest wavelength of the laser, v is the speed of the ship, c is the speed of light,
and ∆λ is the difference between the rest and observed wavelength of the laser (λobserved − λ0 ).
Now for a quick aside about the velocity of the ship...
There is nothing that says that velocity has to be a positive number. When using the doppler shift
equation, we use the convention that objects moving towards us have negative velocity and objects
moving away from us have positive velocity.
If you plug in a negative velocity to the Doppler shift equation, you can see that ∆λ will be negative,
meaning that the observed wavelength will be lower than the rest wavelength. On the other hand, if you
plug in a positive velocity, the observed wavelength will be higher than the rest wavelength.
Because the spaceship is coming towards us, we know the light will be blue-shifted (λobserved < λ0 ).
That means that we should plug in a negative velocity to the equation.
Now that we know we should use −60, 000 km/s for v, we can solve the Doppler shift equation for the
intrinsic wavelength.
λobserved − λ0
v
=
λ0
c
λobserved − λ0
v
× λ0 = × λ0
c
λ0
λ0 v
+ λ0
λobserved − λ0 + λ0 =
c
v
λobserved = λ0
+1
c
v
λobserved
c+1
= λ0 v v
c+ 1
c +1
λobserved
= λ0
v
+
1
c
Plugging in the given values gives us the answer:
5500Å
km/s
−6×104 km/s
3×105 The rest wavelength of the laser is 6875Å.
+1
= 6875Å