Uncertainty in position of a hypothetical subatomic particle is `1 "Å"` and uncertainty in velocity is `(3.3)/(4 pi) xx 10^(5)` m/s then the mass of the particle is approximately ( h = `6.6 xx 10^(-34)` Js) :

Options

A. `2xx10^(-28) kg`
B. `2xx10^(-27) kg`
C. `2xx10^(-29) kg`
D. `4xx10^(-29) kg`

Explanation

To find the mass of the hypothetical subatomic particle, we will use the Heisenberg Uncertainty Principle, which states that the uncertainty in position (Δx) multiplied by the uncertainty in momentum (Δp) is greater than or equal to a constant (h/4π). The formula can be expressed as:

\[ \Delta x \cdot m \cdot \Delta v \geq \frac{h}{4\pi} \]

Where:
- Δx = uncertainty in position
- m = mass of the particle
- Δv = uncertainty in velocity
- h = Planck's constant

### Step 1: Identify the given values
- Uncertainty in position (Δx) = 1 Å = \(1 \times 10^{-10}\) m
- Uncertainty in velocity (Δv) = \(\frac{3.3}{4\pi} \times 10^5\) m/s
- Planck's constant (h) = \(6.6 \times 10^{-34}\) Js

### Step 2: Substitute the values into the uncertainty principle equation
We can rearrange the uncertainty principle equation to solve for mass (m):

\[ m = \frac{h}{4\pi \Delta x \Delta v} \]

### Step 3: Calculate Δv
First, calculate the value of Δv:

\[ \Delta v = \frac{3.3}{4\pi} \times 10^5 \text{ m/s} \]

Calculating \(4\pi\):

\[ 4\pi \approx 12.566 \]

Now, substituting this value:

\[ \Delta v \approx \frac{3.3}{12.566} \times 10^5 \approx 0.262 \times 10^5 \text{ m/s} = 2.62 \times 10^4 \text{ m/s} \]

### Step 4: Substitute Δx and Δv into the equation for mass
Now, substitute Δx and Δv into the mass equation:

\[ m = \frac{6.6 \times 10^{-34}}{4\pi \times (1 \times 10^{-10}) \times (2.62 \times 10^4)} \]

### Step 5: Calculate the denominator
Calculate \(4\pi \times (1 \times 10^{-10}) \times (2.62 \times 10^4)\):

\[ 4\pi \times (1 \times 10^{-10}) \times (2.62 \times 10^4) \approx 12.566 \times 2.62 \times 10^{-6} \approx 3.30 \times 10^{-5} \]

### Step 6: Calculate mass (m)
Now substitute back into the mass equation:

\[ m \approx \frac{6.6 \times 10^{-34}}{3.30 \times 10^{-5}} \approx 2.0 \times 10^{-29} \text{ kg} \]

### Final Answer
The mass of the hypothetical subatomic particle is approximately \(2.0 \times 10^{-29}\) kg.

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