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Physics Question on Dimensional Analysis

The dimension of stopping potential V0V_0 in photoelectric effect in units of Planck's constant 'h', speed of light 'c' and Gravitational constant 'G' and ampere A is :

A

h−2/3c−1/3G4/3A−1h ^{-2/3} c ^{-1/3} G ^{4/3} A ^{-1}

B

h0c5G−1A−1h^{0} c^{5} G^{-1} A^{-1}

C

h2G3/2c1/3A−1h ^{2} G ^{3/2} c^{1/3} A^{-1}

D

h1/3G2/3c1/3A−1h ^{1/3} G ^{2/3} c ^{1/3} A ^{- 1}

Answer

h0c5G−1A−1h^{0} c^{5} G^{-1} A^{-1}

Explanation

Solution

v0=hx cy Gz Awv_{0}=h^{x}\,c^{y}\,G^{z}\,A^{w}
ML2T−2AT=(ML2T−1)x(LT−1)(M−1L3T−2)zAw\frac{ML^{2}T^{-2}}{AT}=\left(ML^{2}T^{-1}\right)^{x}\left(LT^{-1}\right)\left(M^{-1}L^{3}T^{-2}\right)^{z}A^{w}
⇒w=−1\Rightarrow w=-1
(x−z=1)\left(x - z = 1\right)
2x+y+3x=22x + y + 3x = 2
−x−y−2z=−32x=0\frac{-x-y-2z=-3}{2x=0}
x=0x=0
z=−1y=5z=-1 y=5
⇒v0=h0c5G−1A−1\Rightarrow v_{0}=h^{0}c^{5}G^{-1}A^{-1}