The wavelength associated with an electron accelerated through a potential difference of `100 V` is nearly

Options

A. `1000Å`
B. `100Å`
C. `10.5Å`
D. `1.2Å`

Answer

To find the wavelength associated with an electron accelerated through a potential difference of 100 V, we can follow these steps:

### Step 1: Calculate the kinetic energy of the electron
The work done on the electron when it is accelerated through a potential difference \( V \) is given by:
\[ \text{Work done} = eV \]
where \( e \) is the charge of the electron, approximately \( 1.6 \times 10^{-19} \) coulombs.

For \( V = 100 \) V:
\[ \text{Kinetic Energy (KE)} = eV = (1.6 \times 10^{-19} \, \text{C})(100 \, \text{V}) = 1.6 \times 10^{-17} \, \text{J} \]

### Step 2: Relate kinetic energy to momentum
The kinetic energy can also be expressed in terms of momentum \( p \):
\[ KE = \frac{p^2}{2m} \]
where \( m \) is the mass of the electron, approximately \( 9.1 \times 10^{-31} \) kg.

Setting the two expressions for kinetic energy equal to each other:
\[ 1.6 \times 10^{-17} = \frac{p^2}{2 \times 9.1 \times 10^{-31}} \]

### Step 3: Solve for momentum \( p \)
Rearranging the equation to solve for \( p^2 \):
\[ p^2 = 2m \cdot KE = 2 \times (9.1 \times 10^{-31}) \times (1.6 \times 10^{-17}) \]
Calculating this gives:
\[ p^2 = 2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-17} \approx 2.912 \times 10^{-47} \, \text{kg}^2 \text{m}^2/\text{s}^2 \]

### Step 4: Calculate momentum \( p \)
Now take the square root to find \( p \):
\[ p = \sqrt{2.912 \times 10^{-47}} \approx 5.39 \times 10^{-24} \, \text{kg m/s} \]

### Step 5: Calculate the de Broglie wavelength \( \lambda \)
The de Broglie wavelength is given by:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant, approximately \( 6.626 \times 10^{-34} \, \text{J s} \).

Substituting the values:
\[ \lambda = \frac{6.626 \times 10^{-34}}{5.39 \times 10^{-24}} \approx 1.23 \times 10^{-10} \, \text{m} \]

### Step 6: Convert to Angstroms
To convert meters to Angstroms (1 Angstrom = \( 10^{-10} \) m):
\[ \lambda \approx 1.23 \, \text{Å} \]

### Final Answer
The wavelength associated with an electron accelerated through a potential difference of 100 V is approximately \( 1.23 \, \text{Å} \).

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