Question

Two isotopes of a certain element have binding energies that differ by 3.8580 MeV. The isotope with the larger binding energy contains one more neutron than the other isotope. Find the difference in atomic mass between the two isotopes, by taking the energy equivalent of 1 u to be 931.50 MeV. Express your answer in atomic mass units.

Answer 1

Answer:

The difference in atomic mass between the two isotopes is 1.00452329 atomic mass unit.

Explanation:

For an isotope-I (heavier)

Mass of an isotope-I=M

Number of neutrons = n+1

Number of protons = p

$\Delta m_1=((n+1)\times m_n)+(p\times m_p))-M$

For an isotope-II

Mass of an isotope-II=M'

Number of neutrons = n

Number of protons = p

$\Delta m_2=((n)\times m_n)+(p\times m_p))-M'$

Difference in binding energy:

$B.E=\Delta mc^2$ (general binding energy expression)

Binding energy difference between two isotopes:

$\Delta B.E=B.E-B.E'=3.8580 MeV=(\Delta m_1-\Delta m_2)c^2$..(1)

$3.8580 MeV=(\Delta m_1-\Delta m_2)c^2$

$=([((n+1)\times m_n)+(p\times m_p))-M]-[((n)\times m_n)+(p\times m_p))-M'])c^2$

$B.E-B.E'=3.8580 MeV$

$=([((n+1)\times m_n)+(p\times m_p))-M]-[((n)\times m_n)+(p\times m_p))-M'])c^2$

$3.8580 MeV=[1\times m_n-M+M']c^2$

$\frac{3.8580}{931.50} u=m_n-M+M'$

$M-M'=1.008665 u -\frac{3.8580}{931.50} u=1.00452329 u$

The difference in atomic mass between the two isotopes is 1.00452329 atomic mass unit.