I know the answer to 17 is 12.7, the answer to 18 is 127.6 MeV, and the answer t
ID: 2255385 • Letter: I
Question
I know the answer to 17 is 12.7, the answer to 18 is 127.6 MeV, and the answer to 19 is 5.62 MeV. I need the work.
A neutron star is approximately an enormous stable nucleus composed of neutrons. Estimate the radius of a neutron star which has a mass equal to the Sun's mass M sum = 1.99 times 10 30 kg. Determine the binding energy of . Use M n = 1.008665u, M p = 1.007825u and M = 15.994915u. Radon-222 decays by a decay. Write the reaction for this decay and determine the energy which is released. You will need M = 222.017571u, M = 218.008930u, and M = 4.002603u.Explanation / Answer
17) 1.c. A neutron star with twice the mass of the Sun, but the radius of a city (10 km)
Part 1: Data
The procedure is the same, only the numbers are different!
MNS = 1.99 X 1030 kg
RS = 695,500 km
MNS = 1.99 X 1030 kg X 1000 gm / kg = 1.99 X 1033 grams
We must convert the radius into centimeters:
RNS = 695,500 km X (1000m / km) X (100cm / m) =12.7 X 10^30 km
18)
expected mass = 8(1.007825 ) + 8 (1.008665 ) = 16.13192 amu
B) mass defect = expected mass - actual mass
mass defect = 16.13192 - 15.994915
mass defect = 0.137005 amu
so the mass defect is 0.137005 u
C) binding energy = delta m x 931.5 MeV
B.E = 0.137005 x 931.5
B.E = 127.62 MeV
So the Binding energy is 127.62 MeV
19)
mass lost = mass of Rn - mass of alpha - mass of Po
mass lost = 222.017571 - 4.002603 -218.008930
mass lost = 6.038 x 10-3 u
energy released = mass lost x 931.5
energy released = 6.038 x 10-3 x 931.5
energy released = 5.624 MeV