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Determine the binding energy per nucleon of the americium isotope X95244X2952244Am, given the mass of X95244X2952244Am to be 244.06428 u. - Physics

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प्रश्न

Determine the binding energy per nucleon of the americium isotope \[\ce{_95^244Am}\], given the mass of \[\ce{_95^244Am}\] to be 244.06428 u.

संख्यात्मक

उत्तर

Given: Z = 95, N = 244 - 95 = 149, mp = 1.00728 u, mn = 1.00866 u, M = 244.06428 u, 1 u = 931.5 MeV/c2

To find:  B.E. per nucleon `E_B/A =   ?`

The binding energy per nucleon,

`"E"_"B"/"A" = (("Zm"_"p" + "Nm"_"n" - "M")"c"^2)/"A"` 

`= ([95(1.00728) + 149(1.00866) - 244.06428]"uc"^2)/244`

`= ((95.6916 + 150.29034 - 244.06428)/244)(931.5)` MeV nucleon

∴ `E_B/A = ` 7.3209 MeV/nucleon

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अध्याय 15: Structure of Atoms and Nuclei - Exercises [पृष्ठ ३४२]

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बालभारती Physics [English] 12 Standard HSC Maharashtra State Board
अध्याय 15 Structure of Atoms and Nuclei
Exercises | Q 12 | पृष्ठ ३४२

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