An electron is accelerated through a potential difference of 200volts. What is the de-broglie wavelength associated with it? To which part of the electromagnetic spectrum does this value of wavelength correspond?

Explanation

To find the de Broglie wavelength associated with an electron that has been accelerated through a potential difference of 200 volts, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between potential difference and kinetic energy**:
When an electron is accelerated through a potential difference (V), it gains kinetic energy (KE) equal to the work done on it by the electric field. The kinetic energy can be expressed as:
\[ KE = eV \]
where \( e \) is the charge of the electron (approximately \( 1.6 \times 10^{-19} \) coulombs) and \( V \) is the potential difference (200 volts in this case).

2. **Calculate the kinetic energy**:
Substituting the values:
\[ KE = (1.6 \times 10^{-19} \, \text{C}) \times (200 \, \text{V}) = 3.2 \times 10^{-17} \, \text{J} \]

3. **Relate kinetic energy to momentum**:
The kinetic energy can also be expressed in terms of momentum (p) as:
\[ KE = \frac{p^2}{2m} \]
where \( m \) is the mass of the electron (approximately \( 9.1 \times 10^{-31} \, \text{kg} \)). Rearranging this gives us:
\[ p = \sqrt{2m \cdot KE} \]

4. **Substituting the values**:
\[ p = \sqrt{2 \times (9.1 \times 10^{-31} \, \text{kg}) \times (3.2 \times 10^{-17} \, \text{J})} \]
Calculating this gives:
\[ p \approx \sqrt{5.824 \times 10^{-47}} \approx 7.63 \times 10^{-24} \, \text{kg m/s} \]

5. **Use the de Broglie wavelength formula**:
The de Broglie wavelength \( \lambda \) is given by:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant (\( 6.63 \times 10^{-34} \, \text{Js} \)).

6. **Substituting the momentum into the de Broglie wavelength formula**:
\[ \lambda = \frac{6.63 \times 10^{-34} \, \text{Js}}{7.63 \times 10^{-24} \, \text{kg m/s}} \approx 8.69 \times 10^{-11} \, \text{m} \]

7. **Convert to Angstroms**:
Since \( 1 \, \text{Å} = 1 \times 10^{-10} \, \text{m} \):
\[ \lambda \approx 0.87 \, \text{Å} \]

8. **Determine the part of the electromagnetic spectrum**:
The wavelength of \( 0.87 \, \text{Å} \) falls within the gamma rays region of the electromagnetic spectrum, which typically ranges from \( 0.01 \, \text{Å} \) to \( 10 \, \text{Å} \).

### Final Answer:
The de Broglie wavelength associated with the electron is approximately \( 0.87 \, \text{Å} \), which corresponds to the gamma rays part of the electromagnetic spectrum.

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