The dimension of stopping potential $V_{0}$ in photoelectric effect in units of Planck's constant $^{'} h^{'}$ , speed of light $^{'} c^{'}$ , Gravitational constant $^{'} G^{'}$ and ampere $^{'} A^{'}$ is:

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

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h^{2 / 3} G^{5 / 3} c^{1 / 3} A^{- 1}

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

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h^{0} G^{- 1} c^{5} A^{- 1}

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Solution

The correct option is D $h^{0} G^{- 1} c^{5} A^{- 1}$
Let, the dimensional formula for,

$Stopping potential (V_{0}) \propto h^{x} A^{y} G^{z} c^{r}$

Here,
$h \to Planck's constant = [M L^{2} T^{- 1}]$
$A \to current = [A]$
$G \to Gravitational constant = [M^{- 1} L^{3} T^{- 2}]$
and $c \to speed of light = [L T^{- 1}]$

$V_{0} = potential = [M L^{2} T^{- 3} A^{- 1}]$

Using dimensional analysis,

$[M L^{2} T^{- 3} A^{- 1}] = [M L^{2} T^{- 1}]^{x} [A]^{y} [M^{- 1} L^{3} T^{- 2}]^{z} [L T^{- 1}]^{r}$

$[M L^{2} T^{- 3} A^{- 1}] = M^{x - z} L^{2 x + 3 z + r} T^{- x - 2 z - r} A^{y}$
Comparing and solving the dimension of $M, L, T, A$ , we get,

$y = - 1, x = 0, z = - 1, r = 5$
$∴ V_{0} \propto h^{0} A^{- 1} G^{^{- 1}} c^{5}$

Hence, option $(D)$ is correct.

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