JEE PYQ: Units & Measurements - Question ID 395251577d58 (JEE Main 2020)

In the photoelectric effect, the stopping potential $V_0$ is the minimum reverse potential required to halt the most energetic photoelectrons emitted from a metal surface. Its value depends on the incident photon energy $h\nu$ and the work function $\phi$ of the metal, but dimensionally it behaves like an electric potential (energy per unit charge).

The question asks for the dimensional formula of $V_0$ expressed in terms of Planck’s constant $h$, speed of light $c$, gravitational constant $G$, and electric current $A$. We must express $V_0$ as a product of powers of these constants: $V_0 \sim h^a \cdot c^b \cdot G^c \cdot A^d$ and determine the exponents $a, b, c, d$.

Key dimensional formulas:
- $[V_0] = \text{Energy} \cdot \text{Charge}^{-1} = \text{ML}^2\text{T}^{-3}\text{A}^{-1}$
- $[h] = \text{ML}^2\text{T}^{-1}$
- $[c] = \text{LT}^{-1}$
- $[G] = \text{M}^{-1}\text{L}^3\text{T}^{-2}$
- $[A] = \text{A}$

Step 1: Write the dimensional equation
We equate the dimensions of $V_0$ to the product of the dimensions of $h^a c^b G^c A^d$: $\text{ML}^2\text{T}^{-3}\text{A}^{-1} = [h]^a [c]^b [G]^c [A]^d$ Substitute the dimensions of each constant: $\text{ML}^2\text{T}^{-3}\text{A}^{-1} = (\text{ML}^2\text{T}^{-1})^a (\text{LT}^{-1})^b (\text{M}^{-1}\text{L}^3\text{T}^{-2})^c (\text{A})^d$

Step 2: Expand the right-hand side $= \text{M}^{a - c} \cdot \text{L}^{2a + b + 3c} \cdot \text{T}^{-a - b - 2c} \cdot \text{A}^{d}$

Step 3: Equate exponents of M, L, T, A We obtain four equations:
1. Mass: $a - c = 1$
2. Length: $2a + b + 3c = 2$
3. Time: $-a - b - 2c = -3$
4. Current: $d = -1$

Step 4: Solve the system of equations From equation 4: $d = -1$.
From equation 1: $a = 1 + c$.
Substitute $a = 1 + c$ into equation 3: $-(1 + c) - b - 2c = -3 \implies -1 - c - b - 2c = -3 \implies -b - 3c = -2 \implies b + 3c = 2$ Substitute $a = 1 + c$ into equation 2: $2(1 + c) + b + 3c = 2 \implies 2 + 2c + b + 3c = 2 \implies b + 5c = 0$ Now we have: $b + 3c = 2 \quad \text{and} \quad b + 5c = 0$ Subtract the first from the second: $(b + 5c) - (b + 3c) = 0 - 2 \implies 2c = -2 \implies c = -1$ Then $b = 2 - 3c = 2 - 3(-1) = 5$
And $a = 1 + c = 1 + (-1) = 0$

Step 5: Write the final expression We have $a = 0$, $b = 5$, $c = -1$, $d = -1$. Thus: $V_0 \sim h^0 \cdot c^5 \cdot G^{-1} \cdot A^{-1}$ This matches option B.

- Students often confuse the dimensional role of $h$, thinking it must appear in the final expression. However, $h$ cancels out ($a = 0$), which is counterintuitive but mathematically correct.
- Another common mistake is miscounting the exponents of $G$, especially its negative mass dimension. Forgetting that $[G] = \text{M}^{-1}\text{L}^3\text{T}^{-2}$ leads to incorrect signs.
- Always verify the current dimension separately: since $V_0$ has $\text{A}^{-1}$, the exponent of $A$ must be $-1$, which immediately eliminates options A, C, and D if you check this first.

Final Answer: B