Ncert Exercises & Solutions

44. Einstein's mass-energy relation emerging out of his famous theory of relativity relates mass (m) to energy (E) as E= ${mc}^{2}$ , where c is the speed of light in a vacuum. At the nuclear level, the magnitudes of energy are very small. The energy at the nuclear level is usually measured in MeV, where $1 MeV = 1 .6 \times 10^{- 19} J$ ; the masses are measured in the unified atomic mass unit (u) where $1 u = 1 .67 \times 10^{- 27}$ kg.

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

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

Hint: Use E=

{mc}^{2}

Step 1: Find the energy produced by 1 amu.

(a) We know that,
$\begin{array}{l} 1 amu = 1 u = 1 .67 \times 10^{- 27} kg \\ Applying E = {mc}^{2} \\ Energy = E = (1 .67 \times 10^{- 27}) (3 \times 10^{8})^{2} J \\ = 1 .67 \times 9 \times 10^{- 11} J \\ E = \frac{1 .67 \times 9 \times 10^{- 11}}{1 .6 \times 10^{- 19}} MeV = 939 .4 MeV \approx 931 .5 MeV \end{array}$

Step 2: Write the correct formula.

(b) The dimensionally correct relation is:

$1 amu \times c^{2} = 1 u \times c^{2} = 931 .5 MeV$

NEET Information

courses

Company

Botany Questions

Chemistry Questions

Physics Questions

Zoology Questions