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Thomson's Discovery of Electrons

If the total ...

Question

If the total binding energies of $_{1} H^{2},_{2} H e^{4},_{26} F e^{56}$ and $_{92} U^{235}$ nuclei are $2.22, 28.3, 492$ and $1768$ MeV respectively, then identify the most stable nucleus out of the following.

A

_{26} F e^{56}

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B

_{1} H^{2}

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C

_{92} U^{235}

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D

_{2} H e^{4}

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Solution

The correct option is D $_{26} F e^{56}$
Binding energy $/$ nucleon,
$_{1} H^{2} : 2.22 / 2 = 1.11$ MeV
$_{1} H e^{4} : \frac{28.3}{4} = 7.07$ MeV
$_{26} F e^{56} : \frac{492}{56} = 8.78$ MeV
$_{92} U^{235} : \frac{1786}{235} = 7.6$ MeV.
Binding energy $/$ nucleon is maximum for $_{26} F e^{56}$ , hence it is most stable element.

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Similar questions

Q. Binding energies of

_{1} H^{2},_{2} H e^{4},_{26} F e^{56}

_{92} U^{235}

2.2 Mev, 28.4 Mev,

492 Mev

1786 MeV

respectively, which one of the following is more stable?

Q. In the total binding energies of

_{1} H^{2} a n d_{2} H e^{4},_{26} F e^{56}

_{92} U^{235}

nuclei are 2.22, 28.3, 492 and 1786 MeV respectively. Identify the most stable nucleus out of the following

Q. If the total binding energies of

_{1}^{2} H,_{2}^{4} H e,_{26}^{56} F e &_{92}^{235} U

2.22, 28.3, 492

1786

MeV respectively, identify the most stable nucleus of the following.

Q. The amount of energy released in the fusion of

_{1} H^{2}

_{2} H e^{4}

nucleus will be :
[Binding energy per nucleon of

_{1} H^{2} = 1.1 M e V

_{2} H e^{4} = 7 M e V

]

Q. Binding energy per nucleon of

_{1} H^{2}

_{2} H e^{4}

1.1 M e V

7.0 M e V

, respectively. Energy released in the process

_{1} H^{2} +_{1} H^{2} ⟶_{2} H e^{4}

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