Ncert Exercises & Solutions

13.24 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 \({ }_{20}^{41} \mathrm{Ca}\) and \({ }_{13}^{27} \mathrm{Al}\) from the following data:

m(\({ }_{20}^{40} \mathrm{Ca}\)) = 39.962591 u

m(\({ }_{20}^{41} \mathrm{Ca}\)) = 40.962278 u

m(\({ }_{13}^{26} \mathrm{Al}\)) = 25.986895 u

m(\({ }_{13}^{27} \mathrm{Al}\)) = 26.981541 u

Hint: The energy required is given by \(E=\Delta mc^2\).
Step 1: Find the neutron separation energies of the nucleus \({ }_{20}^{41} \mathrm{Ca}\).

F o r {}_{20}^{41}C a : S e p a r a t i o n e n e r g y = 8.363007 M e V

F o r {}_{13}^{27}A l : S e p a r a t i o n e n e r g y = 13.059 M e V

I f a n e u t r o n {}_{0}^{1}n i s r e m o v e d f r o m {}_{20}^{41}C a, t h e c o r r e s p o n d i n g n u c l e a r r e a c t i o n c a n b e w r i t t e n a s :

{}_{20}^{41}C a \to {}_{20}^{40}C a + {}_{0}^{1}n

I t i s g i v e n t h a t :

m ({}_{20}^{40}C a) = 39.962591 u

m ({}_{13}^{27}C a) = 40.962278 u

m ({}_{0}^{1}n) = 1.008665 u

T h e m a s s d e f e c t o f t h i s r e a c t i o n i s g i v e n a s :

∆ m = m ({}_{20}^{40}C a) + ({}_{0}^{1}n) - m ({}_{20}^{41}C a)

= 39.962591 + 1.008665 - 40.962278 = 0.008978 u

B u t 1 u = 931.5 M e V / C^{2}

∴ ∆ m = 0.008978 \times 931.5 M e V / c^{2}

H e n c e, t h e e n e r g y r e q u i r e d f o r n e u t r o n r e m o v a l i s c a l c u l a t e d a s :

E = ∆ m c^{2}

= 0.008978 \times 931.5 = 8.363007 M e V

Step 2: Find the neutron separation energies of the nucleus \({ }_{13}^{27} \mathrm{Al}\).
For \({ }_{13}^{27} \mathrm{Al}\), the neutron removal reaction can be written as:\({ }_{13}^{27} Al \rightarrow{ }_{13}^{26} Al+{ }_{0}^{1} n\)

I t i s g i v e n t h a t :

m ({}_{13}^{26}{Al}) = 25.986895 u

m ({}_{13}^{27}{Al}) = 26.981541 u

T h e m a s s d e f e c t o f t h i s r e a c t i o n i s g i v e n a s :

∆ m = m ({}_{13}^{26}A l) + m ({}_{0}^{1}n) - m ({}_{13}^{27}A l)

= 25.986895 + 1.008665 - 26.981541

= 0.014019 u

= 0.01409 \times 931.5 M e V / c^{2}

H e n c e, t h e e n e r g y r e q u i r e d f o r n e u t r o n r e m o v a l i s c a l c u l a t e d a s :

E = ∆ m c^{2}

= 0.014019 \times 931.5 = 13.059 M e V

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