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# Q Value Consider the nuclear reaction $x + X \rightarrow Y + y$ $(E_x + M_x C^2) + M_X c^2 = (E_Y + M_Y c^2) + (E_y + M_y C^2)$

Where $E_x$ = kinetic energy of the projectile x $M_x c^2$ = rest energy of the projectile x

And similarly $E_Y$ and $M_Y c^2,$ are kinetic energy and restmass energy of product y and $E_Y$ and $M_Y c^2$ are kinetic energy and restmass energy of product y

The target nucleus is assumed to be at rest

The quantity Q represents the difference between the KE of product of reactions and that of reactant $Q = E_Y + E_y - E_x$ …(1)

or $Q = (M_x + M_X - M_Y - M_y)c^2$

= Rest mass energy of reactant – Rest mass energy of product …(2)

The quantity Q is called the energy balance of the reaction or Q-value of the reaction.

If Q is positive then nuclear reaction is called exoergic reaction

If Q is negative then nuclear reaction is called endoergic reaction

The term $E_y$ in equation -(1) represents the recoil energy of the product nucleus. It is usually small and hard to measure but can be eliminated by taking into account the conservation of momentum.

The Q Equation : The analytical relationship between the kinetic energy of the projectile and outgoing particle and the nuclear distintegration energy Q is called the Q-equation

Applying law of conservation of momentum $M_x V_x = M_Y V_Y \cos \phi = M_y V_y \cos \theta$ $M_Y V_Y \sin \phi = M_y V_y \sin \theta$

Eliminating $\phi$ $M_{Y}^{2} V_{Y}^{2} = M_{x}^{2} V_{x}^{2} + M_{y}^{2} V_{y}^{2} - 2M_x M_y V_x V_y \cos \theta$ $2E_Y M_Y = 2E_x M_x + 2E_y M_y - 4(M_x M_y E_x E_y)^{1/2} \cos \theta$ $Q = E_Y + E_y - E_Y = E_x \left ( \dfrac{M_x}{M_Y} - 1 \right ) + E_y \left ( \dfrac{M_y}{M_Y} + 1 \right ) - \dfrac{2}{M_Y} (M_x M_y E_x E_y)^{1/2}\cos \theta$

This is known as Q-value equation

Solution of the Q-equation

Q-value equation can be seen as quadratic equation in $\sqrt{E_y}$. So, solution can be put as $\sqrt{E_y} = V \pm \sqrt{V^2 + W}$

where $V = \sqrt{\dfrac{M_x M_y E_x}{M_Y + M_y}} \cos \theta$ $W = \dfrac{M_Y Q + E_x (M_Y - M_x)}{M_Y + M_y}$

when $\sqrt{E_y}$ is real and positive, the reaction is energetically possible

Endoergic Reaction : Here Q < 0

For example : $_{8}^{16}O (n, d)^{15} \ _7 N$ has Q = -10 MeV $_{6}^{13}C (p, \alpha){5}^{15} B$ has Q = -MeV

(i) For very low energy projectiles $E_x \cong 0, (Q < 0)$ $V^2 + W < 0$

So, $\sqrt{E_y}$ is imaginary and hence, no reaction occurs. $E_x$ is insufficient to start the reaction.

(ii) Threshold Energy : $\mathbf{(E_x)_{thresh}}$

The smallest value of projectile energy (bombarding energy) at which an endoergic reaction can take place is called the threshold energy of that reaction.

Reaction becomes possible when $E_x$ is large enough to make $V^2 + W = 0$

Hence, $E_x = -Q \left [ \dfrac{M_y + M_Y}{M_y + M_Y + M_x - (M_x/M_y/M_Y) \sin^2 \theta} \right ]$ $E_x$ has its minimum possible value at $\theta = 0$ which is $E_{\mathrm{thresh}}$ $(E_x)_{th} = \dfrac{-Q(M_Y + M_y)}{(M_Y + M_y - M_x)}$

The value of $Q / c^2$ is much less than the masses of particles in a nuclear reaction and so $M_X \cong M_Y + M_y - M_x$

So, $(E_x)_{\mathrm{thresh}} = -Q\left (\dfrac{M_x + M_X}{M_x} \right )$

Exoergic Reactions : Here Q > 0

(i) For very Low energy projectiles e.g. thermal neutrons $E_x = 0$

V = 0, $W = \dfrac{M_Y Q}{M_y + M_y}$

So, $E_y = Q \dfrac{M_Y}{M_Y + M_y}$ $E_Y$ is same for all value of $\theta$ $_{6}^{12} C(n, \alpha){4}^{9} Be$