Question

The binding energy per nucleon of ${}_{17}^{35}Cl$ nucleus is

(${}_{17}^{35}Cl=$34.98000 amu, $m_P=$1.007825 amu,$m_n=$1.008665 amu and 1 amu is equivalent to 931MeV)

• A. $4.6MeV$
• B. $5.8MeV$
• C. $6.5MeV$
• D. $8.2MeV$

Answer 1

The correct option is D $8.2MeV$

In $Cl$ there are $17$ protons and $(35-17)$ $=$ 18 neutrons.

$\Delta m=[17\times 1.007825+18\times 1.008665-34.98]amu$

$=0.308995$

$BE=0.308995\times 931.478MeV$

$=287.82MeV$

$\frac{BE}{nucleon}=\frac{287.82}{35}=8.22MeV$