Question

The dimension of stopping potential $V_0$ in photoelectric effect in units of Planck's constant $h$, speed of light $c$ and gravitational constant $G$ and ampere $A$ is

• A. $h^{-2/3}{c}^{-1/3}{G}^{4/3}{A}^{-1}$
• B. $h^{1/3}{G}^{2/3}{c}^{1/3}{A}^{-1}$
• C. $h^2{G}^{3/2}{c}^{1/3}{A}^{-1}$
• D. $=h^0{c}^5{G}^{-1}{A}^{-1}$

Answer 1

Answer: (D)

Let $V_0=(h{)}^a\cdot (c{)}^b\cdot (G{)}^c\cdot (A{)}^d$.....(i)
Then, $\left[V_0\right]=[$ potential $]=\left[\frac{\text{ potential energy }}{\text{ charge }}\right]$
$=\frac{\left[M{L}^2{T}^{-2}\right]}{\left[A{T}^{\prime }\right]}=\left[M{L}^2{T}^{-3}{A}^{-1}\right]$
$[h]=\left[\frac{\text{ Energy }}{\text{ Frequency }}\right]=\frac{\left[M{L}^2{T}^{-2}\right]}{\left[T^{-1}\right]}=\left[M{L}^2{T}^{-1}\right]$
$[c]=[$ Speed $]=\left[L{T}^{-1}\right]$
$[G]=\left[\frac{\text{ Force }\times ({\text{ Distance }}^2}{(\text{ Mass }{)}^2}\right]=\frac{\left[ML{T}^{-2}\right]\left[L^2\right]}{\left[M^2\right]}$
$=\left[M^{-1}{L}^3{T}^{-2}\right]$
Substituting the dimensions of $V_{0}h,C,G$ and $A$ in Eq. (i) and equating dimension on both sides, we get
$\left[M{L}^2{T}^{-3}{A}^{-1}\right]={\left[M{L}^2{T}^{-1}\right]}^a\times {\left[L{T}^{-1}\right]}^b$
$\times {\left[M^{-1}{L}^3{T}^{-2}\right]}^c\times [A{]}^d$
$\Rightarrow a-c=1$.....(ii)
$2a+b+3c=2$......(iii)
$-a-b-2c=-3$.......(iv)
$d=-1$ ......(v)
On solving above equations, we get
$a=0,b=5,c=-1,d=-1$
Substituting these values in Eq. (i), we get
$V_0=h^0\cdot c^5\cdot G^{-1}\cdot A^{-1}$