If the mass of the electron `(9xx10^(-31)` kg) is taken as unit of mass, the radius of the first Bohr orbit `(0.5xx10^(-10)`m) as unit of length and 500 newton as the unit of force, then the unit of time in the new system would be

Options

A. `3 xx10^(-22)` s
B. `15 xx10^(-12)` s
C. `15xx10^(-20)` s
D. `45xx10^(-20)` s

Answer

To find the unit of time in the new system where the mass of the electron, the radius of the first Bohr orbit, and a specific force are defined as the units of mass, length, and force respectively, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given quantities:**
- Mass of the electron (m) = \(9 \times 10^{-31}\) kg (unit of mass)
- Radius of the first Bohr orbit (l) = \(0.5 \times 10^{-10}\) m (unit of length)
- Force (F) = 500 N (unit of force)

2. **Recall the formula for force:**
\[ F = m \cdot a \]
where \(a\) is acceleration.

3. **Express the units of force in terms of mass, length, and time:**
- The dimensional formula for force is:
\[ [F] = [M][L][T^{-2}] \]
where:
- \([M]\) = mass
- \([L]\) = length
- \([T]\) = time

4. **Substituting the new units into the force equation:**
- We have:
\[ 500 = (9 \times 10^{-31}) \cdot a \]
- The acceleration \(a\) can be expressed as:
\[ a = \frac{F}{m} = \frac{500}{9 \times 10^{-31}} \text{ m/s}^2 \]

5. **Express acceleration in terms of the new unit of length and time:**
- Since \(a = \frac{l}{t^2}\), where \(l\) is the unit of length:
\[ a = \frac{0.5 \times 10^{-10}}{t^2} \]

6. **Set the two expressions for acceleration equal to each other:**
\[ \frac{0.5 \times 10^{-10}}{t^2} = \frac{500}{9 \times 10^{-31}} \]

7. **Cross-multiply to solve for \(t^2\):**
\[ 0.5 \times 10^{-10} \cdot 9 \times 10^{-31} = 500 \cdot t^2 \]
\[ t^2 = \frac{0.5 \times 9 \times 10^{-41}}{500} \]

8. **Calculate \(t^2\):**
\[ t^2 = \frac{4.5 \times 10^{-41}}{500} = 9 \times 10^{-42} \]

9. **Take the square root to find \(t\):**
\[ t = \sqrt{9 \times 10^{-42}} = 3 \times 10^{-21} \text{ seconds} \]

### Final Answer:
The unit of time in the new system is \(3 \times 10^{-22}\) seconds.