The binding energies of the nuclei of 42He - 73Li - 126C and 147N are 28 - 52 - 90 - 98 Me V respectively- Which of these is most stable-

The correct option is A ^4_2He The nucleus with higher binding energy will be more stable. #160;We have to calculate the binding energy per nucleus from ...

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Standard VIII

Chemistry

Isotopes and Isotones

The binding e...

Question

The binding energies of the nuclei of $_{2}^{4} H e$ , $_{3}^{7} L i$ , $_{6}^{12} C$ and $_{7}^{14} N$ are $28$ , $52$ , $90$ , $98$ $M e V$ respectively. Which of these is most stable?

_{2}^{4} H e

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_{3}^{7} L i

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_{6}^{12} C

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_{7}^{14} N

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Solution

The correct option is A $_{2}^{4} H e$
The nucleus with higher binding energy will be more stable.

We have to calculate the binding energy per nucleus from the given values of binding energies.

The binding energy of a nucleus is calculated by the following formula :

Binding Energy $= \frac{B i n d i n g E n e r g y p e r N u c l e u s}{M a s s N u m b e r}$

Binding energy per nucleus of $_{2}^{4} H e = \frac{28}{4} = 7$

Binding energy per nuclei of $_{3}^{7} L i = \frac{52}{7} = 7.42$

Binding energy per nuclei of $_{6}^{12} C = \frac{90}{12} = 7.5$

Binding energy per nuclei of $_{7}^{14} N = \frac{98}{14} = 7$

So, the carbon $(_{6}^{12} C)$ nucleus will be more stable because, in carbon, the binding energy per nucleus is maximum $(7.5)$ .

Hence, option (C) is correct.

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The binding energies of the nuclei of 42He , 73Li , 126C and 147N are 28 , 52 , 90 , 98 Me V respectively. Which of these is most stable?

The binding energies of the nuclei of $_{2}^{4} H e$ , $_{3}^{7} L i$ , $_{6}^{12} C$ and $_{7}^{14} N$ are $28$ , $52$ , $90$ , $98$ $M e V$ respectively. Which of these is most stable?