When an electron experiences a potential difference of 150 volt, the wave associated with it will have a wavelenght:

Options

A. `1.0 xx 10^(-5) cm`
B. `1.0 xx 10^(-8) cm`
C. `1.2 xx 10^(-8) cm`
D. `10.0 xx 10^(-8) cm`

Explanation

To find the wavelength associated with an electron that experiences a potential difference of 150 volts, we can use the de Broglie wavelength formula and the relationship between kinetic energy and potential difference. Here’s the step-by-step solution:

### Step 1: Understand the de Broglie wavelength formula
The de Broglie wavelength (\( \lambda \)) is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the electron.

### Step 2: Relate kinetic energy to momentum
The kinetic energy (\( KE \)) of the electron can be expressed in terms of momentum:
\[ KE = \frac{p^2}{2m} \]
From this, we can express momentum \( p \) as:
\[ p = \sqrt{2m \cdot KE} \]

### Step 3: Relate kinetic energy to potential difference
When an electron is accelerated through a potential difference \( V \), its kinetic energy is given by:
\[ KE = eV \]
where \( e \) is the charge of the electron (approximately \( 1.6 \times 10^{-19} \) coulombs) and \( V \) is the potential difference (150 volts in this case).

### Step 4: Substitute the kinetic energy into the momentum formula
Substituting \( KE = eV \) into the momentum equation:
\[ p = \sqrt{2m \cdot eV} \]

### Step 5: Substitute momentum into the de Broglie wavelength formula
Now substituting \( p \) back into the de Broglie wavelength formula:
\[ \lambda = \frac{h}{\sqrt{2m \cdot eV}} \]

### Step 6: Insert values into the equation
Now we can insert the known values:
- \( h = 6.63 \times 10^{-34} \, \text{Js} \)
- \( m = 9.1 \times 10^{-31} \, \text{kg} \)
- \( e = 1.6 \times 10^{-19} \, \text{C} \)
- \( V = 150 \, \text{V} \)

Thus, we have:
\[ \lambda = \frac{6.63 \times 10^{-34}}{\sqrt{2 \cdot 9.1 \times 10^{-31} \cdot 1.6 \times 10^{-19} \cdot 150}} \]

### Step 7: Calculate the denominator
Calculating the denominator:
\[ 2 \cdot 9.1 \times 10^{-31} \cdot 1.6 \times 10^{-19} \cdot 150 = 4.368 \times 10^{-48} \]
Taking the square root:
\[ \sqrt{4.368 \times 10^{-48}} \approx 6.61 \times 10^{-24} \]

### Step 8: Calculate the wavelength
Now substituting back to find \( \lambda \):
\[ \lambda = \frac{6.63 \times 10^{-34}}{6.61 \times 10^{-24}} \approx 1.00 \times 10^{-10} \, \text{m} \]

### Step 9: Convert to centimeters
To convert meters to centimeters:
\[ \lambda \approx 1.00 \times 10^{-10} \, \text{m} = 1.00 \times 10^{-8} \, \text{cm} \]

### Final Answer
The wavelength associated with the electron is approximately:
\[ \lambda \approx 1.00 \times 10^{-8} \, \text{cm} \]

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