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. \(h^\frac{-2}{3}G^{\frac{4}{3}}c^{\frac{-1}{3}}A^{-1}\)


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 \[[hG] = [c]^5\]


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.