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The binding enrgy of .(17)^(35)Cl nucleu...

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

Text Solution

Verified by Experts

The correct Answer is:
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
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