JEE Advance - Physics (2016 - Paper 2 Offline - No. 8)
The measured masses of the neutron, $$_1^1H$$, $$_7^{15}N$$ and $$_8^{15}O$$ are 1.008665u, 1.007825u, 15.000109u and 15.003065u, respectively. Given that the radii of both the $$_7^{15}N$$ and $$_8^{15}O$$ nuclei are same, 1 u = 931.5 MeV/c2 (c is the speed of light) and e2/(4$$\pi$$$${{\varepsilon _0}}$$) = 1.44 MeV fm. Assuming that the difference between the binding energies of $$_7^{15}N$$ and $$_8^{15}O$$ is purely due to the electrostatic energy, the radius of either of the nuclei is (1 fm = 10$$-$$15 m)
Explanation
The binding energies of $$_7^{15}N$$ and $$_8^{15}O$$ are given by
$$B{E_N} = \Delta {m_N}{c^2} = (8{m_n} + 7{m_p} - {M_N}){c^2}$$
$$ = (8 \times 1.008665 + 7 \times 1.007825 - 15.000109 \times 931.5$$
= 115.49 MeV,
$$B{E_O} = \Delta {m_O}{c^2} = (7{m_n} + 8{m_p} - {M_O}){c^2}$$
$$ = (7 \times 1.008665 + 8 \times 1.007825 - 15.003065 \times 931.5$$
= 111.95 MeV.
The difference in binding energies of $$_7^{15}N$$ and $$_8^{15}O$$ is
$$\Delta BE = B{E_N} - B{E_O} = 115.49 - 111.95$$
= 3.54 MeV. ........ (1)
The electrostatic energies of $$_7^{15}N$$ and $$_8^{15}O$$ are given by
$${E_N} = {3 \over 5}{{Z(Z - 1){e^2}} \over {4\pi { \in _0}R}} = {3 \over 5}{{7(7 - 1)1.44} \over R}$$
$$ = {{36.288} \over R}$$ MeV-fm
$${E_O} = {3 \over 5}{{8(8 - 1)1.44} \over R} = {{48.384} \over R}$$ MeV-fm.
The difference in electrostatic energies of $$_7^{15}N$$ and $$_8^{15}O$$ is
$$\Delta E = {E_O} - {E_N} = (12.096/R)$$ ...... (2)
Since, $$\Delta E = \Delta BE$$, equations (1) and (2) give $$R = 12.096/3.54 = 3.42$$ fm.
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