Question

Find Binding energy of an $\alpha -$particle in $MeV$?
$[m_{proton}=1.007825amu,m_{neutron}=1.008665amu,m_{helium}=4.002800amu]$

• A. $28.097eV$
• B. $28.097MeV$
• C. $38.097eV$
• D. $48.097MeV$

Answer 1

Answer: (B)

$28.097MeV$

$\begin{array}{cc}\text{Given}-& \ast m_{\text{proton}}=1.007825\text{amu}\\ & \ast m_{\text{neutron}}=1.008665\text{amu}\\ & \ast m_{\text{Helium}}=4.002800\text{amu.}\end{array}$

$\text{For the reaction below-}$

$2\cdot {}_1^{1}H+2{\cdot }_0^{1}H⟶{}_2^{4}He$

$\begin{array}{c}{\text{We get ,Binding energy of}}_2^{4}He\\ \Rightarrow BE=\Delta m\cdot c^2=[m_{\text{Helium}}-2m_{\text{Proton}}-2m_{\text{neutron}}]c^2\\ \begin{array}{cc}BE=& \left(0.03018\right)\text{amu}\times c^2=0.03018\times 931MeV\\ & =28.097MeV\end{array}\end{array}$