# Question 1

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:

1. $$h^2Gc^{-5}A^{-1}$$
2. $$h^\frac{1}{3}G^{\frac{2}{3}}c^{\frac{1}{3}}A^{-1}$$
3. $$h^\frac{2}{3}G^{\frac{1}{3}}c^{\frac{5}{3}}A^{-1}$$
4. $$h^\frac{-2}{3}G^{\frac{4}{3}}c^{\frac{-1}{3}}A^{-1}$$

## Concepts

In the original problem that appeared in IIT Mains, all options were wrong and full marks were awarded to each student. This problem has a very basic difficulty. Not only all options were incorrect, this question does not have a unique answer as $$h, c, G, A$$ do not form an independent set of dimensions like $$M, L, T, A$$ would form. It is easily verified that dimensionally we have $[hG] = [c]^5$

## Solution

The options have been modified such that it contains one correct option. Let us formally see how to solve this problem.

#### Step 1

Express the dimensions of $$V_0$$ in terms of $$h, c, G, A$$ with unknown powers $$x_h, x_c, x_G, x_A$$ respectively. $[V_0] = [h]^{x_h}[G]^{x_G}[c]^{x_c}[A]^{x_A}$

#### Step 2

Write the dimensions of all the quantities envolved in terms of $$M, L, T, A$$. $\begin{eqnarray} [V_0] &=& [ML^2T^{-3}A^{-1}] \\ [h] &=& [ML^2T^{-3}] \\ [G] &=& [M^{-1}L^3T^{-2}] \\ [c] &=& [LT^{-1}] \end{eqnarray}$

#### Step 3

Substitute these expressions into Equation in Step 1, $\begin{eqnarray} [ML^2T^{-3}A^{-1}] &=& [ML^2T^{-3}]^{x_h} [M^{-1}L^3T^{-2}]^{x_G} [LT^{-1}]^{x_c} [A]^{x_A} \\ &=& [M^{x_h-x_G}L^{2x_h+3x_G+x_c}T^{-3x_h-2x_G-x_c}A^{x_A}] \end{eqnarray}$

#### Step 4

Equate the powers of like terms to get a system of equations $\begin{eqnarray} x_h-x_G &=& 1\\ 2x_h+3x_G+x_c &=& 2\\ -3x_h-2x_G-x_c &=& 3\\ x_A &=& -1 \end{eqnarray}$

##### Step 5

Solve the system of equations. This system does not have a unique solution. A generic solution can be expressed as \left.\begin{aligned} x_h &= x_G + 1\\ x_c &= -5x_G\\ x_A &= -1 \end{aligned}\right\rbrace \rightarrow \left\{\begin{aligned} x_G &= 1\\ x_h &= 2\\ x_c &= -5\\ x_A &= -1 \end{aligned}\right.

That is, $$[V_0] = [h^2Gc^{-5}A^{-1}]$$. Hence, the correct choice is Option 1.