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Nuclear Fusion

Determine the...

Question

Determine the energy released during the fusion of $_{1} H^{1}$ to form a $_{2} {H e}^{4}$ nucleus, if the total mass of the products is $4.001506 u$ , mass of proton is $1.0078 u$ and mass of neutron is $1.00878 u$ .

27.67 M e V

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27.67 J

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127.67 J

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127.67 M e V

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Solution

The correct option is A $27.67 M e V$
$2_{0} n^{1} +_{1} H^{1} +_{1} H^{1} \to_{2} {H e}^{4} △ m = 2 m (_{0} n^{1}) + 2 m (_{1} H^{1}) - m (_{2} {H e}^{4}) = 2 \times 1.00878 + 2 \times 1.0078 - 4.001506 = 2.01756 + 2.0156 - 4.001506 = 0.0316 u E = △ m c^{2}, 1 u = 931 M e V / c^{2} = 0.0316 \times 931 E = 27 M e V$
So, (A) is correct option.

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Similar questions

_{2} {H e}^{4}

1.007 u, 1.009 u, 4.002 u

(_{1} D^{2})

D (D, p) T

(_{2} {H e}^{4})

T (D, n)_{2} {H e}^{4}

(_{1} D^{2}) = 2.014102

_{1} T^{3} = 3.016049 a . m . u

_{2} {H e}^{4} = 4.002603 a . m . u

_{1} H^{1} = 1.007825 a . m . u

_{0} n^{1} = 1.008665 a . m . u

1.0078 u

1.0087 u

4.00388

= 1.00757

= 1.00893

_{3} {L i}^{6} +_{0} n^{1} \to_{2} {H e}^{4} +_{1} H^{3}

_{3} {L i}^{6}

= 6.015126 a m u

_{1} H^{3}

= 3.016049 a m u

_{2} {H e}^{4}

= 4.002604 a m u

_{0} n^{1}

= 1.008665 a m u

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