Stopping potential depends on planks constant (h), current (I), universal gravitational constant (G) and speed of light (C) choose the correct option for the dimension of stopping potential (V).

Options

A. `hI^-1G^1C^5`
B. `h^-1I^1G^-1C^6`
C. `h^0I^1G^1C^6`
D. `h^0I^-1G^-1C^5`

Explanation

To find the dimensional formula for stopping potential (V) in terms of Planck's constant (h), current (I), universal gravitational constant (G), and speed of light (c), we can follow these steps:

### Step 1: Understand the definition of stopping potential
Stopping potential is defined as the potential energy per unit charge. The dimensional formula for energy is given by:
\[ [E] = [M][L^2][T^{-2}] \]
And for charge, it can be expressed in terms of current (I) and time (T):
\[ [Q] = [I][T] \]
Thus, the dimensional formula for stopping potential (V) can be expressed as:
\[ [V] = \frac{[E]}{[Q]} = \frac{[M][L^2][T^{-2}]}{[I][T]} = [M][L^2][T^{-3}][I^{-1}] \]

### Step 2: Write the relationship for stopping potential
According to the problem, stopping potential (V) depends on Planck's constant (h), current (I), gravitational constant (G), and speed of light (c). We can express this relationship as:
\[ V \propto h^a \cdot c^b \cdot I^p \cdot G^q \]
Where \(a\), \(b\), \(p\), and \(q\) are the powers to be determined.

### Step 3: Write the dimensional formulas for each variable
- For Planck's constant (h):
\[ [h] = [M][L^2][T^{-1}][I^{-1}] \]
- For speed of light (c):
\[ [c] = [L][T^{-1}] \]
- For current (I):
\[ [I] = [I] \]
- For gravitational constant (G):
\[ [G] = [M^{-1}][L^3][T^{-2}] \]

### Step 4: Substitute the dimensional formulas into the relationship
Substituting the dimensional formulas into the equation gives us:
\[ [M][L^2][T^{-3}][I^{-1}] = [M^{a-1}][L^{2a+b+3q}][T^{-a-b-2q}][I^{-p}] \]

### Step 5: Equate the powers of M, L, T, and I
From the equation, we can equate the powers of M, L, T, and I:
1. For M: \(1 = a - q\)
2. For L: \(2 = 2a + b + 3q\)
3. For T: \(-3 = -a - b - 2q\)
4. For I: \(-1 = -p\)

### Step 6: Solve the equations
From the equations:
1. From \(1 = a - q\), we get \(a = 1 + q\).
2. Substitute \(a\) in the second equation:
\[ 2 = 2(1 + q) + b + 3q \implies 2 = 2 + 2q + b + 3q \implies b + 5q = 0 \implies b = -5q \]
3. Substitute \(a\) and \(b\) in the third equation:
\[ -3 = -(1 + q) - (-5q) - 2q \implies -3 = -1 - q + 5q - 2q \implies -3 = -1 + 2q \implies 2q = -2 \implies q = -1 \]
4. Using \(q = -1\) in \(a = 1 + q\):
\[ a = 1 - 1 = 0 \]
5. Using \(q = -1\) in \(b = -5q\):
\[ b = -5(-1) = 5 \]
6. From \(-1 = -p\), we get \(p = 1\).

### Step 7: Write the final expression for stopping potential
Substituting the values of \(a\), \(b\), \(p\), and \(q\) back into the relationship gives us:
\[ V = k \cdot h^0 \cdot c^5 \cdot I^{-1} \cdot G^{-1} \]
Thus, the dimensional formula for stopping potential is:
\[ V \propto c^5 \cdot I^{-1} \cdot G^{-1} \]

### Conclusion
The correct option for the dimensions of stopping potential is:
\[ h^0, c^5, I^{-1}, G^{-1} \]