Question

The threshold energy (in MeV) for the following nuclear reaction to proceed is
${}_2^{4}He+{}_7^{14}N\to {}_8^{17}O+{}_1^{1}H$
Atomic mass of ${}_2^{4}He=4.00260amu$
Atomic mass of ${}_7^{14}N=14.00307amu$
Atomic mass of ${}_8^{17}O=16.99913amu$
Atomic mass of ${}_1^{1}H=1.00783amu$

Answer 1

Answer: 1.54

$E_{\text{thereshold }}=(-Q)+(-Q)\left(\frac{m_{He}}{m_{N_2}}\right)$
$Q=\left(m_{He}+m_{N_2}-m_O-m_H\right)c^2$