The dimension of stopping potential mathrm{V} | vedclass

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Question

$\mathrm{v}_{0}=\mathrm{h}^{\mathrm{x}} \mathrm{c}^{\mathrm{y}} \mathrm{G}^{\mathrm{z}} \mathrm{A}^{\mathrm{w}}$
$\frac{\mathrm{ML}^{2} \mathrm{T}^{-

Vedclass · Accepted Answer

$\mathrm{h}^{0} \mathrm{c}^{5} \mathrm{G}^{-1} \mathrm{A}^{-1}$

Vedclass · Answer

$\mathrm{v}_{0}=\mathrm{h}^{\mathrm{x}} \mathrm{c}^{\mathrm{y}} \mathrm{G}^{\mathrm{z}} \mathrm{A}^{\mathrm{w}}$
$\frac{\mathrm{ML}^{2} \mathrm{T}^{-2}}{\mathrm{AT}}=\left(\mathrm{ML}^{2} \mathrm{T}^{-1}ight)^{\mathrm{x}}\left(\mathrm{LT}^{-1}ight)^{\mathrm{y}}\left(\mathrm{M}^{-1} \mathrm{L}^{3} \mathrm{T}^{-2}ight)^{\mathrm{z}} \mathrm{A}^{\mathrm{w}}$
$\Rightarrow \mathrm{w}=-1$
$(\mathrm{x}-\mathrm{z}=1)$
$(\mathrm{x}-\mathrm{z}=1)$
$-\mathrm{x}-\mathrm{x}=1-2 \mathrm{z}=2$
$-\mathrm{x}-\mathrm{y}-2 \mathrm{z}=-3$
$\mathrm{x}-\frac{1}{2} \mathrm{x}=0$
$\mathrm{x}=0$
$\mathrm{x}=-1$
$2 	imes 0+\mathrm{y}+3 \mathrm{x}-1=2$
$y=5$
$\Rightarrow \mathrm{v}_{0}=\mathrm{h}^{0} \mathrm{c}^{5} \mathrm{G}^{-1} \mathrm{A}^{-1}$

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

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