Question

The dimension of stopping potential ${𝑉}_0$ in photoelectric effect in units of Planck’s constant$\left(h\right)$, speed of light$\left(c\right)$, and gravitational constant $\left(G\right)$ and Ampere $\left(A\right)$ is

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

Answer 1

The correct option is C

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

Step 1: Given that:

The dimensions of the stopping potential in terms of Planck's constant, Gravitational constant, speed of light and Ampere is,

$\mathrm{dimensions}\mathrm{of}\left[V\right]={\left(h\right)}^a{\left(I\right)}^b{\left(G\right)}^c{\left(c\right)}^d$ where a,b,c, and d are the integers.

Note that unit of stopping potential is${𝑉}_0$volt.

Know that

Step 2: Planck's Constant Dimensional Formula:

We know that,

Energy of photon is given by,

$\begin{array}{rcl}E& =& \frac{h}{v}\\ h& =& \frac{E}{v}\end{array}$

Dimension of $E=\left[M{L}^2{T}^{-2}\right],\nu =\left[T^{-1}\right]$

Thus, Planck’s constant$\left(h\right)=\left[M{L}^2{T}^{-1}\right]$

Step 3: Gravitational Dimensional Formula:

Gravitational force acting between two objects of masses $m_{1}and{m}_2$​ separated by distance $r$,

$\begin{array}{rcl}F& =& \frac{G{m}_1{m}_2}{r^2}\\ G& =& \frac{F{r}^2}{m_1{m}_2}\end{array}$

Now substituting the dimensional formula for $F,mandr$

$\begin{array}{rcl}G& =& \frac{\left[ML{T}^{-2}\right]\left[L^2\right]}{{\left[M\right]}^2}\\ & =& \left[M^{-1}{L}^3{T}^{-2}\right]\end{array}$

The actual dimensions of stopping potential is $\left[V\right]=M{L}^2{T}^{-3}{A}^{-1}$

Step 4: Calculate the physical quantity of the dimension

Therefore,

$$\begin{gathered} M{L}^2{T}^{-3}{A}^{-1}={\left(M{L}^2{T}^{-1}\right)}^a{\left(A\right)}^b{\left(M^{-1}{L}^3{T}^{-2}\right)}^c{\left(L{T}^{-1}\right)}^d \\ M{L}^2{T}^{-3}{A}^{-1}=M^{a-c}{L}^{2a+3c+d}{T}^{-a-2c-d}{A}^b \end{gathered}$$

Now compare the right-hand side with the left-hand side,

$$\begin{gathered} a-c=1 \\ 2a+3c+d=2 \\ -a-2c-d=-3 \\ b=-1 \end{gathered}$$

On solving,

$$\begin{gathered} c=-1 \\ a=0 \\ d=5 \\ b=-1 \end{gathered}$$

Therefore the solution is $\left[V\right]=\left[{\left(h\right)}^0{\left(A\right)}^{-1}{\left(G\right)}^{-1}{\left(c\right)}^5\right]$.

Hence, the correct option is C.