Question

A nuclear reaction is given as
$$p+ \ ^{15}N\rightarrow \ _Z^AX+n$$
a) Find $$A, Z$$ and identify the nucleus $$X$$.
b) Find the $$Q$$ value of the reaction.
c) If the proton were to collide with the $$^{15}N$$ at rest, find the minimum $$KE$$ needed by the proton to initiate the above reaction

Answer 1

a) The nuclear reaction will be $$_1^1p+ \ _7^{15}N=\ _Z^AX+ \ _0^1n$$

Now, $$1+15=A+1 \Rightarrow A=15$$ and $$1+7=Z+0 \Rightarrow Z=8$$

The nucleus would be isotope of oxygen i.e $$_8^{15}O$$

b) The nuclear reaction will be $$_1^1p+\ _7^{15}N=\ _Z^AX+\ _0^1n$$

Now, $$1+15=A+1 \Rightarrow A=15$$ and $$1+7=Z+0 \Rightarrow Z=8$$

The nucleus would be isotope of oxygen i.e $$_8^{15}O$$

The atomic mass of proton $$=1.00727$$ amu ; atomic mass of neutron $$=1.00866$$ amu ; atomic mass of $$^{15}O=15.9994$$ amu and atomic mass of $$^{15}N=15.000108$$ amu

Sum of masses of reactants $$=1.00727+15.000108=16.0074$$ amu

Sum of masses of products $$=15.9994+1.00866=17.0081$$ amu

The Q-value $$=$$Mass difference $$\Delta M=17.0081-16.0074=1.0007$$ amu$$=1.0007\times 931 MeV=931.65 MeV$$

c) K.E. of proton to initiate the reaction is equal to Q value of reaction $$=\triangle m c^2$$

$$\triangle m={ m }_{ (p) }+{ m }_{ (N) }-{ m }_{ (X) }-{ m }_{ (n) }\\ =1.007825+15.000-\ ^{ 15 }{ O }-n\\ =1.007825+15-15.0031-1.008665\\ =-0.00394\\ $$

The K.E needed by the proton to initiate the reaction $$Q=-0.00394\times 931MeV\\ =-3.66MeV$$

Negative sign signifies that this energy is supplied to initiate the reaction.