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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.

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प्रश्न

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 at nuclear level is usually measured in MeV, where 1 MeV= 1.6 × 10–13 J; the masses are measured in unified atomic mass unit (u) where 1u = 1.67 × 10–27 kg.

  1. Show that the energy equivalent of 1 u is 931.5 MeV.
  2. 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.
दीर्घउत्तर
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उत्तर

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 = 1 u = 1.67 × 10–27 kg

Applying E = mc2

Energy E = (1.67 × 10–27)(3 × 108)2 J = 1.67 × 9 × 10–11 J

E = `(1.67 xx 9 xx 10^-11)/(1.6 xx 10^-13)` MeV

= 939.4 MeV ≈ 931.5 MeV

b. As E = mc2 ⇒ m = `E/c^2`

According to this, 1 u = `(931.5  MeV)/c^2`

Hence the dimensionally correct relation 1 amu × c2 = 1u × c2 = 931.5 MeV.

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पाठ 2: Units and Measurements - Exercises [पृष्ठ १२]

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एनसीईआरटी एक्झांप्लर Physics Exemplar [English] Class 11
पाठ 2 Units and Measurements
Exercises | Q 2.44 | पृष्ठ १२

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