Dimension of Energy, E=[ML2T−2]
Dimension of gravitational constant, G=[M−1L3T−2]
Dimension of Planck's constant, h=[ML2T−1]
Dimension of speed of light, c=[M0LT−1]
Given :
E=Gphqcr
∴ [ML2T−2]=[M−1L3T−2]p× [ML2T−1]q ×[M0LT−1]r
⟹ [ML2T−2]=[M(−p+q)L(3p+2q+r)T(−2p−q−r)]
Equating both sides, we get:
−p+q=1 .............(1)
3p+2q+r=2 .............(2)
−2p−q−r=−2 .............(3)
Adding (2) and (3), we get: p+q=0 .......(4)
Solving (1) and (4),
⟹p=−12 and q=12
Now from (2),
3×(−12)+2×12+r=2
⟹r=52