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