1. Find the dimension of stopping potential

Stopping potential is just electric potential difference:

$V_0 = \frac{\text{work}}{\text{charge}}$

So,

$[V_0] = \frac{[\text{energy}]}{[\text{charge}]}$

Energy has dimension:

$[E] = M L^2 T^{-2}$

Charge has dimension:

$[Q] = A T$

Hence,

$[V_0] = M L^2 T^{-3} A^{-1}$

2. Write dimensions of the given constants

• Planck's constant:

$[h] = M L^2 T^{-1}$

• Speed of light:

$[c] = L T^{-1}$

• Gravitational constant:

From

$F = \frac{G m_1 m_2}{r^2}$

we get

$[G] = M^{-1} L^3 T^{-2}$

• Ampere:

$[A] = A$

3. Assume

$[V_0] = [h]^x [c]^y [G]^z [A]^{-1}$

Substitute dimensions:

$M L^2 T^{-3} A^{-1} = (M L^2 T^{-1})^x (L T^{-1})^y (M^{-1} L^3 T^{-2})^z A^{-1}$

Collect powers:

$M^{x-z} L^{2x+y+3z} T^{-x-y-2z} A^{-1}$

Now compare with

$M^1 L^2 T^{-3} A^{-1}$

So we get equations:

1. $x-z = 1$
2. $2x+y+3z = 2$
3. $x+y+2z = 3$

4. Solve the equations

From (1):

$x = 1+z$

Put into (3):

$(1+z)+y+2z = 3$ $y+3z = 2$ $y = 2-3z$

Put into (2):

$2(1+z) + (2-3z) + 3z = 2$ $2+2z+2-3z+3z = 2$ $4+2z = 2$ $2z = -2$ $z = -1$

Then,

$x = 1+(-1)=0$

and

$y = 2-3(-1)=5$

Thus,

$[V_0] = h^0 c^5 G^{-1} A^{-1}$

5. Match with options

This corresponds to:

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

6. Comparison with stored correct answer

Stored correct answer: B

Derived answer: B

They agree.