Question

What is the binding energy per nucleon of ${}^{\mathbf{262}}\mathbf{Bh}$? The mass of the atom is 262.1231u.

Short answer

The binding energy per nucleon of is $7.31\mathrm{MeV}\mathrm{per}\mathrm{nucleon}$.

Step-by-step solution

The given data

1. The given atom is ${}^{262}\mathrm{Bh}$with atomic number Z = 107 and mass number A = 262.
2. The mass of the atom,$m_{Bh}=262.1231u$
3. The atomic mass of the hydrogen,$m_H=1.007825u$
4. The atomic mass unit of neutron,$m_n=1.008665u$

Solve for the binding energy per nucleon of the atom:  

The binding energy per nucleon of an atom is as follows:

$\mathbf{∆}{\mathit{E}}_{\mathbf{b}\mathbf{e}\mathbf{/}\mathbf{n}\mathbf{u}\mathbf{c}\mathbf{l}\mathbf{e}\mathbf{o}\mathbf{n}}\mathbf{=}\frac{\left[Z{m}_H+\left(A-Z\right)m_n-m_{atom}\right]{\mathbf{c}}^{\mathbf{2}}}{\mathbf{A}}$ …… (i)

Calculate the value of binding energy per nucleon

Using the given data in equation (i), determine the binding energy per nucleon of bohrium atom${}^{262}\mathrm{Bh}$ as follows:

$$\begin{gathered} ∆{\mathrm{E}}_{\mathrm{be}/\mathrm{nucleon}}=\frac{\left[\left(107\right)\left(1.007825\mathrm{u}\right)+\left(262-107\right)\left(1.008665\mathrm{u}\right)-\left(262.1231\mathrm{u}\right)\right]\left(931.5{\displaystyle \frac{\mathrm{MeV}}{\mathrm{u}}}\right)}{262} \\ =7.31\mathrm{MeV}\mathrm{per}\mathrm{nucleon} \end{gathered}$$

Hence, the value of energy per nucleon is 7.31 MeV per nucleon.