Question

Calculate the binding energy per nucleon for a 157N715N nucleus. The mass of the neutral atom of 157N715N is 15.000109 uu, the mass of the neutral atom of 11H11H is 1.007825 uu and the mass of neutron is 1.008665

Answer 1

Answer:

$1.85\cdot 10^{-11}J$

Explanation:

The binding energy of a nucleus is given by

$\Delta E=c^2 \Delta m$ (1)

where

$c=3.0\cdot 10^8 m/s$ is the speed of light

$\Delta m$ is the mass defect, which is the difference between the sum of the masses of the individual nucleons and the total mass of the nucleus.

Here the mass of the nucleus is

$m(^{15}_7N) = 15.000109 u$

While the mass of the nucleons is:

$m(p)=1.007825 u$ (mass of the proton)

$m(n)=1.008665u$ (mass of the neutron)

The nucleus $^{15}_7N$ containes 7 protons (atomic number) and 15 nucleons (mass number), which means that the number of neutrons is

$n=15-7=8$

So the mass defect is:

$\Delta m=(7m(p)+8m(n))-m(^{15}_7N)=\\(7\cdot 1.007825+8\cdot 1.008665)-15.000109)=0.123986u$

1 atomic mass unit is

$1 u = 1.66054\cdot 10^{-27}kg$

So the mass defect in kilograms is

$\Delta m=(0.123986)(1.66054\cdot 10^{-27})=2.059\cdot 10^{-28}kg$

Finally, we can use eq(1) to find the binding energy:

$\Delta E=(3\cdot 10^8)^2 (2.059\cdot 10^{-28})=1.85\cdot 10^{-11}J$