Question

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

Answer 1

Given:
Stopping potential ,\((V_0)\propto h^xA^yG^zc^r...(i))\)

Where,
ℎ= plank's constant \(=\left [ ML^2T^{-1} \right ]\)
𝐼= current \(=\left [ A \right ]\)
𝐺= Gravitational constant \(=\left [M^{-1}L^3T^{-2} \right ]\)
𝑐= speed of light \(=\left [ LT^{-2} \right ]\)
𝑉= potential \(=\left [ ML^2T^{-3}A^{-1} \right ]\)

By putting all the dimensions in equation (𝑖) we get,

\(\left [ ML^2T^{-3}A^{-1} \right ]=\left [ ML^2T^{-1} \right ]^x\left [ A \right ]^y\left [ M^{-1}L^3T^{-2} \right ]^s\left [ LT^{-1} \right ]^r\)

\(\left [ ML^2T^{-3}A^{-1} \right ]=\left [ M \right ]^{x-z}\left [ L \right ]^{2x+3z+r}\left [ T \right ]^{-x-2z-r}\left [ A \right ]^y\)

Comparing the exponents on both sides,
\(1=x-z...(ii)\)
\(1=2x+3z+r...(iii)\)
\(-3=-x-2z-r...(iv)\)
\(-1=y...(v)\)

By solving all the equations, we get
\(y=-1,~x=0,~z=-1,~r=5\)

Therefore, by putting the values in equation (𝑖) we get,
\(v\propto h^0A^{-1}G^{-1}c^5\)

Final Answer: (𝑐)