Question

Einstein’s mass energy relation emerging out of his famous theory of relativity relates mass (m) to energy E as E = mc² , where c is speed of light in vacuum. At the nuclear level, the magnitudes of energy are very small. The energy are very small. The energy at nuclear level is usually measured in MeV, where 1 MeV = 1.6 × 10^-27 kg.

(a) Show that the energy equivalent of 1 u is 931.5 Mev.

(b) A student writes the relation as 1 u = 931.5 MeV. The teacher points out that the relation is dimensionally incorrect. Write the correct relation.

Answer 1

(a) We can apply Einstein’s mass energy relation in this problem, E = mc², to calculate the energy equivalent of the given mass.

Here,

1 amu = 1u = 1.67 × 10^-27 kg

Applying E = mc²,

E = (1.67 × 10^-27)(3 × 10^-8)²J

= 1.67×9 ×10^-11J

or, E = \(\frac{1.67\times9\times10^{-11}}{1.6\times10^{-13}}MeV\)

= 939.3 MeV

≈ 931.5 MeV

(b) As E = mc²

⇒ \(m=\frac{E}{c^2}\)

According to this 1 u = \(\frac{931.5meV}{c^2}\)

Hence, the dimensionally correct relation is 

1 amu × c² = 1u × c²

= 931.5 MeV