Question

calculate the amount of heat energy inside the nucleus of the sun

Answer 1

You are driving and inadvertently run a red traffic light and a police officer pulls you over. You decide to tell the officer that you thought the light was green. You explain the Doppler shift of light to the officer and how since you were moving toward the traffic light the wavelengths of light were getting blueshifted. How fast would you have to have been driving for this to be true? Give your answer in both km/s and miles/hr. Should the officer give you a ticket?

(NOTE: use the wavelength of the red light to be 7000 Å and that of green light to be 5000 Å)

The Doppler shift formula is /0 = v/c.  = 5000 Å - 7000 Å = - 2000 Å (the minus sign indicates a blueshift). So,

/0 = 2000Å/7000Å = 0.29 
(I've dropped the minus sign, because we know we are talking about a blueshift now.)

v/c = 0.29 ---> v = 0.29c
v = 0.29(3x105 km/s) = 8.6 x 104 km/s

8.6 x 104 km/s * 0.6214 miles/km * 3600 s/hr = 1.9 x 108 miles/hr = 190 million miles per hour !!!

Yes, you should definitely get a ticket, for speeding!

E = mc-who?

Using Einstein's famous equation E = mc2, show that when 4 protons fuse into 1 helium nucleus via the proton-proton cycle, about 4.14 x 10-12 Joules of energy is released. (The mass of one individual proton is about 1.6725 x 10-27kg; the mass of one helium nucleus is about 6.644 x 10-27 kg.)

The difference in the masses of the Helium nucleus and the 4 protons is 4.6 x10-29kg.

(4 x 1.6725 x 10-27kg) - 6.644 x 10-27kg = 4.6 x10-29kg

E = (4.6 x10-29kg)x( 9.0x1016 m2/s2) = 4.14x10-12 Joules

(recall one Joule is a kgm2/s2)

The total luminosity (energy radiated per second) of the sun is about 3.85 x 1026 J/s. How many nuclei of helium are being created per second?

The number of He nuclei produced each second corresponds to the number of fusion reactions occuring (as each reaction produces one He nucleus). Divide the energy output of the Sun by the energy of one fusion reaction to find out how many reactions per second are required to power it, and we'll have our Helium count as well.

(3.85x1026 J/s)/(4.14x10-12 J/reaction) = 9.29 x1037 reactions (Helium nuclei)/s.

What percentage of the hydrogen mass is converted into energy via fusion?

The proportional change in mass is the actual change 4.6x10-29kg (note that we can write it as .046 x10-27 kg so that it more easily cancels), divided by the original mass of 4x1.6725x10-27kg = 6.690 x10-27kg.

(.046 x10-27kg)/(6.690 x10-27kg) = 0.0068

which expressed as a percentage is 0.68 % ( 0.7 % for those of us who round)

How much mass in the Sun is being converted into energy every second?

We can solve this by dividing the given Luminosity by c2(A) or by multiplying the values we calculated for reactions per second and mass-loss per reaction (B). Either approach gives us 4.27x109kg.

Answer 2

Ans: According to Einstein's mass energy relation :
E = mc^2
E=9 × 10^6 J