The binding enrgy of `._(17)^(35)Cl` nucleus is 298 MeV. Find the atomic mass. Given, mass of a proton `(m_(P))=1.007825` amu, mass of a neutron `(m_(

Question

The binding enrgy of `._(17)^(35)Cl` nucleus is 298 MeV. Find the atomic mass. Given, mass of a proton `(m_(P))=1.007825` amu, mass of a neutron `(m_(

The binding enrgy of `._(17)^(35)Cl` nucleus is 298 MeV. Find the atomic mass. Given, mass of a proton `(m_(P))=1.007825` amu, mass of a neutron `(m_(n))=1.008665` amu.
A. `24.9` amu
B. `34.9` amu
C. `54.9` amu
D. `35. 289` amu

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - B
m = `17 m_(rho) + 18 m_(a)`
= `17 xx 1.007825 + 18 xx 1.008665`
= `35. 289`
`Delta m = ( 298 MeV)/(931.2 MeV// "amu") = 0.3200` amu
`therefore` Atomic mass = m - `Delta`m
= `35. 289 - 0.32`
= `34.969` amu

4 views Answered 2023-02-05 15:29