Find the minimum kinetic energy (in eV ) of an α-particle to cause the reaction 14 N (α, p)17 O. The masses of 14 N , 4 He , 1 H and 17 O are respectively 14.00307 u, 4.00260 u , 1.007 .83 u and 16.99913 u.

Question

Find the minimum kinetic energy (in eV ) of an α-particle to cause the reaction 14 N (α, p)17 O. The masses of 14 N , 4 He , 1 H and 17 O are respectively 14.00307 u, 4.00260 u , 1.007 .83 u and 16.99913 u. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Since the masses are given in atomic mass units, it is easiest to proceed by finding the mass difference between reactants and products in the same units and then multiplying by

931.5 M e V / u

. Thus, we have

Q = (14.00307 u + 4.00260 u

- 1.00783 u - 16.99913 u) (931.5 \frac{M e V}{u})

= - 1.20 M e V

Q

value is negative. It means reaction is endothermic.
So, the minimum kinetic energy of

α

-particle to initiate this reaction would be,

K_{m i n} = ∣ Q ∣ (\frac{m _{α}}{m _{N}} + 1)

= (1.20) (\frac{4.00260}{14.00307} + 1)

= 1.54 M e V

Q. Find the minimum kinetic energy (in eV ) of an α-particle to cause the reaction 14N(α,p)17O. The masses of 14N,4He,1H and 17O are respectively 14.00307u, 4.00260u,1.007.83u and 16.99913u.

Q. Find the minimum kinetic energy (in $e V$ ) of an $α$ -particle to cause the reaction $^{14} N (α, p)^{17} O$ . The masses of $^{14} N,^{4} He,^{1} H$ and $^{17} O$ are respectively $14.00307 u$ , $4.00260 u, 1.007.83 u$ and $16.99913 u$ .