If an electron having kinetic energy `2 eV` is accelerated through the potential difference of `2` volt. Then calculate the wavelength associated with the electron

Explanation

To solve the problem of calculating the wavelength associated with an electron that has a kinetic energy of 2 eV and is accelerated through a potential difference of 2 volts, we can follow these steps:

### Step 1: Understand the relationship between kinetic energy and potential difference
When an electron is accelerated through a potential difference (V), the kinetic energy (KE) gained by the electron is given by the equation:
\[ KE = eV \]
where \( e \) is the charge of the electron (approximately \( 1.6 \times 10^{-19} \) coulombs) and \( V \) is the potential difference in volts.

### Step 2: Calculate the total kinetic energy of the electron
In this case, the electron initially has a kinetic energy of 2 eV. After being accelerated through a potential difference of 2 volts, the total kinetic energy becomes:
\[ KE_{total} = KE_{initial} + eV \]
\[ KE_{total} = 2 \, \text{eV} + 2 \, \text{eV} = 4 \, \text{eV} \]

### Step 3: Convert kinetic energy from eV to joules
To use the kinetic energy in calculations involving wavelength, we need to convert it from electron volts to joules. The conversion factor is:
\[ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \]
Thus,
\[ KE_{total} = 4 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 6.4 \times 10^{-19} \, \text{J} \]

### Step 4: Use the kinetic energy to find the 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 \( 9.11 \times 10^{-31} \) kg). Rearranging this gives:
\[ p = \sqrt{2m \cdot KE} \]
Substituting the values:
\[ p = \sqrt{2 \times (9.11 \times 10^{-31} \, \text{kg}) \times (6.4 \times 10^{-19} \, \text{J})} \]

### Step 5: Calculate the momentum
Calculating the value:
\[ p = \sqrt{2 \times 9.11 \times 10^{-31} \times 6.4 \times 10^{-19}} \]
\[ p \approx \sqrt{1.165 \times 10^{-48}} \]
\[ p \approx 1.08 \times 10^{-24} \, \text{kg m/s} \]

### Step 6: Calculate the wavelength using de Broglie's equation
According to de Broglie's hypothesis, the wavelength (\( \lambda \)) associated with a particle is given by:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant (\( 6.63 \times 10^{-34} \, \text{Js} \)). Substituting the values:
\[ \lambda = \frac{6.63 \times 10^{-34}}{1.08 \times 10^{-24}} \]

### Step 7: Calculate the wavelength
Calculating the value:
\[ \lambda \approx 6.14 \times 10^{-10} \, \text{m} \]
To convert this to angstroms (1 angstrom = \( 10^{-10} \, \text{m} \)):
\[ \lambda \approx 61.4 \, \text{angstroms} \]

### Final Answer
The wavelength associated with the electron is approximately \( 61.4 \, \text{angstroms} \).

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