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How Much Energy is Released in the Following Reaction : 7li + P → α + α. Atomic Mass of 7li = 7.0160 U and that of 4he = 4.0026 U. - Physics

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प्रश्न

How much energy is released in the following reaction : 7Li + p → α + α.
Atomic mass of 7Li = 7.0160 u and that of 4He = 4.0026 u.

(Use Mass of proton mp = 1.007276 u, Mass of `""_1^1"H"` atom = 1.007825 u, Mass of neutron mn = 1.008665 u, Mass of electron = 0.0005486 u ≈ 511 keV/c2,1 u = 931 MeV/c2.)

योग
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उत्तर

Given:-
Mass of 7Li = 7.0160 u
Mass of 4He = 4.0026 u.

Reaction :-

`L_i^7 + p → alpha + alpha + E` ,

Energy release (E) is given by

`E = [m(""^7L_i) + (m_p) - 2 xx m ("^4H_e)] c^2`

`= [(7.0160  "u" + 1.007276  "u") - 2(4.0026  "u")]c^2`

`= (8.023273  "u" - 8.0052  "u") c^2`

`= 0.018076 xx 931  "MeV"`

`= 16.83  "MeV"`

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अध्याय 24: The Nucleus - Exercises [पृष्ठ ४४२]

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एचसी वर्मा Concepts of Physics Vol. 2 [English] Class 11 and 12
अध्याय 24 The Nucleus
Exercises | Q 4 | पृष्ठ ४४२

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