Find the binding energy per nucleon of 79197 Au if its atomic mass is 196.96 u. Where m p =1.007277 u m n =1.00866 u (in MeV / nucleus )

Question

Find the binding energy per nucleon of 79197 Au if its atomic mass is 196.96 u. Where m p =1.007277 u m n =1.00866 u (in MeV / nucleus ) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Δ m =79 m p +118 M n - (atomic mass -79 m e ) =79 × 1.007277+118 × 1.00866+79× 0.00055-196.96000 =1.680213 amu Hence ( text B.E. / A )=(1.6802 × 931.5/197) MeV Ans. =7.94 MeV / nucleon

Q. Find the binding energy per nucleon of $_{79}^{197} A u$ if its atomic mass is $196.96 u$ . Where $m_{p} = 1.007277 u & m_{n} = 1.00866 u$ (in $M e V /$ nucleus )

$Δ m = 79 m_{p} + 118 M_{n} -$ (atomic mass $- 79 m_{e}$ )
$= 79 \times 1.007277 + 118 \times 1.00866 + 79 \times 0.00055 - 196.96000$
$= 1.680213 am u$
Hence $\frac{B.E.}{A} = \frac{1.6802 \times 931.5}{197} M e V$
Ans. $= 7.94 M e V /$ nucleon

Q. Find the binding energy per nucleon of 79197​Au if its atomic mass is 196.96u. Where mp​=1.007277u&mn​=1.00866u (in MeV/ nucleus )

Δm=79mp​+118Mn​− (atomic mass −79me​ ) =79×1.007277+118×1.00866+79×0.00055−196.96000 =1.680213amu Hence A B.E. ​=1971.6802×931.5​MeV Ans. =7.94MeV/ nucleon

Q. Find the binding energy per nucleon of $_{79}^{197} A u$ if its atomic mass is $196.96 u$ . Where $m_{p} = 1.007277 u & m_{n} = 1.00866 u$ (in $M e V /$ nucleus )

$Δ m = 79 m_{p} + 118 M_{n} -$ (atomic mass $- 79 m_{e}$ )
$= 79 \times 1.007277 + 118 \times 1.00866 + 79 \times 0.00055 - 196.96000$
$= 1.680213 am u$
Hence $\frac{B.E.}{A} = \frac{1.6802 \times 931.5}{197} M e V$
Ans. $= 7.94 M e V /$ nucleon