uncertainty in the measurement of energy when an electron is excited to state in which it can stay for 1 nano second is

Options

A. `5.25 xx10^(-26)` j
B. `6.6 xx10^(-34)`j
C. `1 xx10^(-9)` j
D. `1.22 xx 10^(-12)` j

Answer

To solve the problem of finding the uncertainty in the measurement of energy when an electron is excited to a state in which it can stay for 1 nanosecond, we will use the Heisenberg Uncertainty Principle. The principle states that the uncertainty in energy (ΔE) multiplied by the uncertainty in time (Δt) is approximately equal to a constant, which can be expressed as:

\[ \Delta E \cdot \Delta t \approx \frac{h}{4\pi} \]

where \(h\) is Planck's constant.

### Step-by-Step Solution:

1. **Identify the given values**:
- Time (\(Δt\)) = 1 nanosecond = \(1 \times 10^{-9}\) seconds
- Planck's constant (\(h\)) = \(6.626 \times 10^{-34}\) Joule seconds (Note: The value used in the video is slightly different, but we will use the more precise value here).

2. **Substitute the values into the uncertainty principle equation**:
\[ \Delta E \cdot \Delta t \approx \frac{h}{4\pi} \]

Plugging in the values:
\[ \Delta E \cdot (1 \times 10^{-9}) \approx \frac{6.626 \times 10^{-34}}{4 \times 3.14} \]

3. **Calculate the denominator**:
\[ 4 \times 3.14 = 12.56 \]

4. **Calculate \(\frac{h}{4\pi}\)**:
\[ \frac{6.626 \times 10^{-34}}{12.56} \approx 5.28 \times 10^{-35} \text{ Joules} \]

5. **Now solve for \(\Delta E\)**:
\[ \Delta E \approx \frac{5.28 \times 10^{-35}}{1 \times 10^{-9}} = 5.28 \times 10^{-26} \text{ Joules} \]

6. **Final Result**:
The uncertainty in the measurement of energy when an electron is excited to a state in which it can stay for 1 nanosecond is approximately:
\[ \Delta E \approx 5.28 \times 10^{-26} \text{ Joules} \]