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Question: Given energy, E=g¹hmCn, where G= universal gravitational constant, h= Planck's constant, C= velocity...

Given energy, E=g¹hmCn, where G= universal gravitational constant, h= Planck's constant, C= velocity of light. Calculate the values of 1, m and n?

Answer

l = 1/8, m = 9/8, and n = -5/8

Explanation

Solution

The energy E is given by the relation E=GlhmCnE = G^l h^m C^n. To find the values of l, m, and n, we equate the dimensions on both sides of the equation.

The dimensions of the quantities are:

  • Energy (E): [ML2T2][ML^2T^{-2}]
  • Universal Gravitational Constant (G): [M1L3T2][M^{-1}L^3T^{-2}]
  • Planck's constant (h): [ML2T1][ML^2T^{-1}]
  • Velocity of light (C): [LT1][LT^{-1}]

Substituting these dimensions into the given relation: [ML2T2]=([M1L3T2])l([ML2T1])m([LT1])n[ML^2T^{-2}] = ([M^{-1}L^3T^{-2}])^l ([ML^2T^{-1}])^m ([LT^{-1}])^n [ML2T2]=[Ml+mL3l+2m+nT2lmn][ML^2T^{-2}] = [M^{-l+m} L^{3l+2m+n} T^{-2l-m-n}]

Equating the powers of M, L, and T on both sides: For M: 1=l+m1 = -l + m (1) For L: 2=3l+2m+n2 = 3l + 2m + n (2) For T: 2=2lmn-2 = -2l - m - n (3)

From (1), m=l+1m = l + 1. Substitute m into (3): 2=2l(l+1)n-2 = -2l - (l+1) - n 2=3l1n-2 = -3l - 1 - n n=3l1n = 3l - 1 (4)

Substitute m and n into (2): 2=3l+2(l+1)+(3l1)2 = 3l + 2(l+1) + (3l-1) 2=3l+2l+2+3l12 = 3l + 2l + 2 + 3l - 1 2=8l+12 = 8l + 1 1=8l    l=181 = 8l \implies l = \frac{1}{8}

Now find m and n: m=l+1=18+1=98m = l + 1 = \frac{1}{8} + 1 = \frac{9}{8} n=3l1=3(18)1=381=58n = 3l - 1 = 3\left(\frac{1}{8}\right) - 1 = \frac{3}{8} - 1 = -\frac{5}{8}

The values are l=18l = \frac{1}{8}, m=98m = \frac{9}{8}, and n=58n = -\frac{5}{8}.