from the following combinations of physical constants (expressed through their as usual symbols) the only combination, that would have the same value in different systems of units, is:

Options

A. `(ch)/(2piepsilon_(o)^(2))`
B. `(e^(2))/(2pi epsilon_(o)Gm_(e)^(2))`
C. `(mu_(o)epsilon_(o))/(c^(2))(G)/(he^(2))`
D. `(2pi sqrt(mu_(o)epsilon_(o))/(ce^(2))(h)/(G)`

Answer

To solve the problem of identifying which combination of physical constants has the same value in different systems of units, we need to analyze each option for dimensional consistency. The only combination that will yield a dimensionless quantity will have the same value regardless of the unit system used.

Let's evaluate the options step by step:

### Step 1: Analyze Option A
**Combination:** \( \frac{ch}{2\pi \epsilon_0^2} \)

- **Constants:**
- \( c \) (speed of light): dimensions \( [L T^{-1}] \)
- \( h \) (Planck's constant): dimensions \( [M L^2 T^{-1}] \)
- \( \epsilon_0 \) (permittivity of free space): dimensions \( [M^{-1} L^{-3} T^4 A^2] \)

**Calculating Dimensions:**
\[ \text{Dimensions of } \epsilon_0^2 = (M^{-1} L^{-3} T^4 A^2)^2 = M^{-2} L^{-6} T^8 A^4 \]
Now substituting into the expression:
\[ \frac{ch}{2\pi \epsilon_0^2} = \frac{[L T^{-1}][M L^2 T^{-1}]}{M^{-2} L^{-6} T^8 A^4} \]
Calculating the dimensions:
\[ = \frac{M L^3 T^{-2}}{M^{-2} L^{-6} T^8 A^4} = M^{3} L^{9} T^{-10} A^{-4} \]
This is not dimensionless.

**Conclusion:** Option A is not dimensionless.

### Step 2: Analyze Option B
**Combination:** \( \frac{e^2}{2\pi \epsilon_0 g m_e^2} \)

- **Constants:**
- \( e \) (charge): dimensions \( [M^{1/2} L^{3/2} T^{-1}] \)
- \( g \) (gravitational constant): dimensions \( [M^{-1} L^3 T^{-2}] \)
- \( m_e \) (mass of electron): dimensions \( [M] \)

**Calculating Dimensions:**
\[ e^2 = (M^{1/2} L^{3/2} T^{-1})^2 = M L^3 T^{-2} \]
Now substituting into the expression:
\[ \frac{e^2}{2\pi \epsilon_0 g m_e^2} = \frac{M L^3 T^{-2}}{M^{-1} L^3 T^{-2} \cdot M^2} \]
Calculating the dimensions:
\[ = \frac{M L^3 T^{-2}}{M^{1} L^3 T^{-2}} = M^{0} L^{0} T^{0} = 1 \]
This is dimensionless.

**Conclusion:** Option B is dimensionless.

### Step 3: Analyze Option C
**Combination:** \( \mu_0 \epsilon_0 g \frac{h}{c^2} e^2 \)

- **Constants:**
- \( \mu_0 \) (permeability of free space): dimensions \( [M L T^{-2} A^{-2}] \)
- \( \epsilon_0 \): dimensions \( [M^{-1} L^{-3} T^4 A^2] \)
- \( g \): dimensions \( [M^{-1} L^3 T^{-2}] \)
- \( h \): dimensions \( [M L^2 T^{-1}] \)
- \( c \): dimensions \( [L T^{-1}] \)

**Calculating Dimensions:**
The combined dimensions will be complex and will not simplify to a dimensionless quantity upon calculation.

**Conclusion:** Option C is not dimensionless.

### Step 4: Analyze Option D
**Combination:** \( \sqrt{\frac{\mu_0 \epsilon_0}{g}} \)

- **Constants:** Same as above.

**Calculating Dimensions:**
This will also yield a complex dimensional analysis that does not simplify to a dimensionless quantity.

**Conclusion:** Option D is not dimensionless.

### Final Answer:
The only combination that is dimensionless is **Option B**: \( \frac{e^2}{2\pi \epsilon_0 g m_e^2} \).