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Additional Exercises · 13.24

Q.The neutron separation energy is defined as the energy required to remove a neutron from the nucleus. Obtain the neutron separation energies of the nuclei 2041Ca^{41}_{20}\text{Ca} and 1327Al^{27}_{13}\text{Al} from the following data:
m(2040Ca)=39.962591m(^{40}_{20}\text{Ca}) = 39.962591 u
m(2041Ca)=40.962278m(^{41}_{20}\text{Ca}) = 40.962278 u
m(1326Al)=25.986895m(^{26}_{13}\text{Al}) = 25.986895 u
m(1327Al)=26.981541m(^{27}_{13}\text{Al}) = 26.981541 u

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✓ Free question

Sn=[m(A−1,Z)+mn−m(A,Z)]c2S_n = [m(A-1,Z) + m_n - m(A,Z)]c^2 for each nucleus. Result: Ca-41 needs ≈8.36\approx 8.36 MeV to remove a neutron; Al-27 needs ≈13.06\approx 13.06 MeV — noticeably more, since Al-27 is more tightly bound.

The neutron separation energy is the energy required for the process ZAX→ZA−1X+n^{A}_{Z}X \rightarrow {}^{A-1}_{Z}X + n, so:

Sn=[m(ZA−1X)+mn−m(ZAX)]c2S_n = \left[m(^{A-1}_{Z}X) + m_n - m(^{A}_{Z}X)\right]c^2

For 2041Ca→2040Ca+n^{41}_{20}\text{Ca} \rightarrow {}^{40}_{20}\text{Ca} + n

Sn=[39.962591+1.008665−40.962278]×931.5 MeVS_n = [39.962591 + 1.008665 - 40.962278] \times 931.5\ \text{MeV}

=[40.971256−40.962278]×931.5= [40.971256 - 40.962278] \times 931.5

=0.008978×931.5=8.363 MeV= 0.008978 \times 931.5 = 8.363\ \text{MeV}

For 1327Al→1326Al+n^{27}_{13}\text{Al} \rightarrow {}^{26}_{13}\text{Al} + n

Sn=[25.986895+1.008665−26.981541]×931.5S_n = [25.986895 + 1.008665 - 26.981541] \times 931.5

=[26.995560−26.981541]×931.5= [26.995560 - 26.981541] \times 931.5

=0.014019×931.5=13.058 MeV= 0.014019 \times 931.5 = 13.058\ \text{MeV}

Al-27's neutron separation energy is notably higher than Ca-41's — Ca-41 has one neutron beyond the especially stable, doubly-magic Ca-40 core (Z=20Z=20, N=20N=20), so that 'extra' 21st neutron is only loosely bound (a well-known nuclear shell-structure effect), while Al-27 has no such magic-number neighbor and is more typically tightly bound.

✓Final answer

Sn(41Ca)≈8.36 MeVS_n(^{41}\text{Ca}) \approx \boxed{8.36\ \text{MeV}}, Sn(27Al)≈13.06 MeVS_n(^{27}\text{Al}) \approx \boxed{13.06\ \text{MeV}}

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