Binding energies of $_{1} H^{2},_{2} H e^{4},_{26} F e^{56}$ and $_{92} U^{235}$ nuclei are $2.2 Mev, 28.4 Mev,$ $492 Mev$ and $1786 MeV$ respectively, which one of the following is more stable?

_{1} H^{2}

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_{2} H e^{4}

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_{26} F e^{56}

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_{92} U^{235}

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Solution

The correct option is C $_{26} F e^{56}$
Stability is measured by B.E per nucleon,

For $_{1} H^{2} : E_{b n} = \frac{2.2}{2} = 1.1 MeV$

For $_{2} H e^{4} : E_{b n} = \frac{28.4}{4} = 7.1 MeV$

For $_{26} F e^{56} : E_{b n} = \frac{492}{56} = 8.8 MeV$

For $_{92} U^{235} : E_{b n} = \frac{1786}{235} = 7.6 MeV$

As we know that more stable nucleus has more binding energy per nucleon.

Thus, $_{26} F e^{56}$ is the most stable.

Hence, $(C)$ is the correct answer.

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Similar questions

Binding energies $_{1} H^{2},_{2} H^{4},_{26} F e^{56}, a n d_{92} U^{235}$ of nuclei are 2.22MeV, 28.4MeV, 492MeV and 1768 MeV respectively. Which one of the following is more stable?

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Binding energies of 1H2, 2He4, 26Fe56 and 92U235 nuclei are 2.2 Mev, 28.4 Mev, 492 Mev and 1786 MeV respectively, which one of the following is more stable?

Binding energies of $_{1} H^{2},_{2} H e^{4},_{26} F e^{56}$ and $_{92} U^{235}$ nuclei are $2.2 Mev, 28.4 Mev,$ $492 Mev$ and $1786 MeV$ respectively, which one of the following is more stable?