Question: An alpha particle with kinetic energy7.70 MeV collides with an ¹⁴Nnucleus at rest, and the two transform into an ¹⁷Onucleus and a proton. The proton is emitted at 90° to the direction of the incident alpha particle and has a kinetic energy of 4.44. MeV The masses of the various particles are alpha particle$\mathbf{4}\mathbf{.}\mathbf{00260}\mathbf{}\mathit{u}{\mathbf{,}}^{\mathbf{14}}\mathit{N}\mathbf{,}\mathbf{14}\mathbf{.}\mathbf{00307}\mathbf{}\mathit{u}$; proton, $\mathbf{1}\mathbf{.}\mathbf{007825}\mathbf{}\mathit{u}$; and$\mathit{u}{\mathbf{,}}^{\mathbf{17}}\mathit{O}\mathbf{,}\mathbf{16}\mathbf{.}\mathbf{99914}\mathbf{}\mathit{u}$In,MeV what are (a) the kinetic energy of the oxygen nucleus and (b) the of the reaction? (Hint:The speeds of the particles are much less than c.)

• The kinetic energy of the alpha particle $K_{\alpha }=7.7MeV$.
• The kinetic energy of the proton is $K_p=4.44MeV$.
• Mass of alpha particle is $m_{\alpha }=4.00260u$.
• Mass of the nitrogen nuclei is$m_N=14.00307u$
• Mass of the proton is$m_p=1.007825u$

Mass of the oxygen nucleus is $m_0=16.99914u$.

The energy an object has as a result of motion is known as kinetic energy. An object's kinetic energy will remain constant as long as it is moving at the same speed. The velocity and mass of an object are used to compute its kinetic energy.

Note: In all calculation, classical approximations are used because it is given as a hint in the question that velocities are much less than the speed of light.

Let’s assume that the alpha particle is travelingalong positive x-axis. Applying conservation of momentum is the x-direction

$m_{\alpha }{u}_{\alpha x}+m_N{u}_{Nx}=m_O{u}_{Ox}+m_p{u}_{px}$

Where, the subscript in the velocity terms represents the velocity along x-direction. The velocity of nitrogen is zero as its stationery and velocity of the proton is zero along the x-axis as its motion is purely along the y-axis.

data-custom-editor="chemistry" $$\begin{gathered} m_{\alpha }{u}_{\alpha x}+0=m_O{u}_{Ox}+0 \\ {u}_{Ox}=\frac{m_{\alpha }{u}_{\alpha x}}{m_O} \end{gathered}$$ …(1)

Now to determine the velocity component of the oxygen nucleus along the x-axis, we need the velocity of the alpha particle. We are given the kinetic energy of the particle and therefore velocity can be determined using the kinetic energy relation

$$\begin{gathered} K_{\alpha }=7.7MeV \\ \frac{1}{2}{m}_{\alpha }{u}_{\alpha x}^2=7.7MeV \\ {u}_{ax}=\sqrt{\frac{2\times \left(7.7\times {10}^{6}eV\right)\left(1.6\times {10}^{-19}J/eV\right)}{4.00260u\left(1.67\times {10}^{-27}Kg/u\right)}} \\ =1.92\times {10}^{7}m/s \end{gathered}$$

Inserting this value in equation (1),

$$\begin{gathered} u_{Ox}=\frac{m_{\alpha }{u}_{\alpha x}}{m_O} \\ =\frac{\left(4.00260u\right)\left(1.92\times {10}^{7}m/s\right)}{16.99914u} \\ =0.45\times {10}^{7}m/s \end{gathered}$$

Now applying the conservation momentum in y-direction,

$m_{\alpha }{u}_{\alpha y}+m_N{u}_{Ny}=m_O{u}_{Oy}+m_p{u}_{py}$

where, the subscript in the velocity terms represent the velocity along y-direction. The velocity of alpha particle is zero as it moves along purely x-axis. The velocity of nitrogen nucleus is zero as it is stationary. And assuming the proton which is moving at to alpha’s direction of motion is moving along purely y-axis.

$$\begin{gathered} 0+0=m_O{u}_{Oy}+m_p{u}_{py} \\ {u}_{Oy}=-\frac{m_p{u}_{py}}{m_O} \end{gathered}$$ …(2)

Now to determine velocity component of oxygen nucleus along y-axis, we need velocity of the proton. We are given kinetic energy of the proton and therefore velocity can be determined using the kinetic energy relation

$$\begin{gathered} K_p=4.44MeV \\ \frac{1}{2}{m}_p{u}_{py}^2=4.44MeV \\ {u}_{ax}=\sqrt{\frac{2\times \left(4.44\times {10}^{6}eV\right)\left(1.6\times {10}^{-19}Je/V\right)}{1.007825u\left(1.67\times {10}^{-27}kg/u\right)}} \\ =2.89\times {10}^{7}m/s \end{gathered}$$

Inserting this value in equation (2),

$$\begin{gathered} u_{Oy}=-\frac{m_p{u}_{py}}{m_O} \\ =-\frac{\left(1.007825u\right)\left(2.89\times {10}^7\right)}{16.99914u} \\ =0.17\times {10}^{7}m/s \end{gathered}$$

Now find the resultant velocity of ¹⁷O,

$$\begin{gathered} u_{res}=\sqrt{u_{Ox}^2+u_{Oy}^2} \\ =\sqrt{{\left(0.45\times {10}^{7}m/s\right)}^2+{\left(0.17\times {10}^{7}m/s\right)}^2} \\ =0.48\times {10}^{7}m/s \end{gathered}$$

Now we can finally find the kinetic energy of oxygen nucleus,

$$\begin{gathered} K_O=\frac{1}{2}{m}_O{u}_{res}^2 \\ =0.5\left(16.99914u\times 1.67\times {10}^{-27}Kg/u\right){\left(0.48\times {10}^{7}m/s\right)}^2 \\ =3.48\times {10}^{-13}J \\ =2.17MeV \end{gathered}$$

In a nuclear or chemical reaction, total energy change in the system is called the Q value. It is defined as

$Q=\left(M_i-M_f\right)c^2$

Where, $M_i$is the sum of masses of reactants and $M_f$is the sum of masses of products.

Therefore, the Q value for the given nuclear reaction is

$$\begin{gathered} Q=\left[\left(m_{\alpha }+m_N\right)-\left(m_O+m_p\right)\right]\left(1.67\times {10}^{-27}kg/u\right){\left(3\times {10}^{8}m/s\right)}^2 \\ =\left[\left(4.00260u+14.00307u\right)-\left(16.99914u+1.007825u\right)\right]\left(1.67\times {10}^{-27}kg/u\right){\left(3\times {10}^{8}m/s\right)}^2 \\ =-1.03MeV \end{gathered}$$

The negative Q value indicates that total mass of the system increases, as a result we can say that, Q amount of kinetic energy is converted into mass.