Question

The atomic mass of ${}_7{N^{15}}$ is 15.000108 amu and that of ${}_8{O^{16}}$ is 15.994915 amu. The minimum energy required to remove the least tightly bound proton is (mass of proton is 1.007825amu)

(A) 0.0130181 MeV

(B) 12.13 MeV

(C) 13.018 MeV

(D) 12.13 MeV

Answer 1

Hint Write the reaction taking place and find out the mass in amu. Now convert the mass into eV by using the conversion factor to obtain the minimum energy to remove the proton.

Complete step-by-step solution

The given elements are undergoing a radioactive decay reaction where the emission of protons is taking place. Here, the oxygen atom with mass 15.994915 amu is the reactant and the nitrogen with mass 15.000108 amu with a proton of mass 1.007825 amu are the products.

${}_8{O^{16}} \to {}_7{N^{15}} + p$

Before you calculate the energy we need to calculate $\Delta m$ which is given by,

$\Delta m$= mass of products – mass of reactants.

$\Delta m$= (mass of ${}_7{N^{15}}$+ mass of p) – mass of ${}_8{O^{16}}$

$$\Delta m = (15.000108 + 1.007825) - 15.994915 \\\ \Delta m = 0.013018amu \\\$$

The mass now obtained is in terms of amu. So, we know that 1 amu = 931 MeV

So,

$
E = 0.013018 \times 931 \\

E = 12.13MeV \\

$

Hence, the minimum energy is 12.13 MeV and the correct option is B

Note According to Einstein, mass and energy are inter convertible. The Einstein mass energy relationship is given by $E = m{c^2}$. From here when m=1 amu and c=$3 \times {10^8}m{s^{ - 1}}$; E= 931 MeV.