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.
(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.
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.
(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.
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1 Answer
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This is a long answer type question as classified in NCERT Exemplar
(a) 1 amu = 1.67 10-27kg
And energy E=mc2= (1.67 10-27) (3 )2J
E= 939.4MeV
(b) So dimensionally correct relation will be E=amu c2
= 1uc2
= 931.5MeV
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According to question, we can write
Total moles of gas = n = nOxygen + nOxygen =
Volume of gas = 12 × 22.4 litre = 268.8 litre = 2.688 × 105 cm3
Viscosity = Pascal . Second
=
= x = 1, x + 2y = -1, -x + z = 1
y = -1, Z = 0
Viscosity = [P1A-1T0]
1 msD = 1mm
10 vsD = 9msD
1vsD = 0.9 MsD
L.C. = 1MSD – 1VSD = 1 – 0.9 = 0.1 mm
Zero error = 4LC = 0.4 mm
Reading = MSR + VSR + correction
= 4.1 cm + 6 * .01 cm + (-0.04 cm) = (4.1 + 0.06 – 0.04) cm
= 4.12 cm = 412 * 10-2 cm
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