The de Broglie wavelength `lamda` of an electron accelerated through a potential V in volts is

Options

A. `(1.227)/(sqrt(V))nm`
B. `(0.1227)/(sqrt(V))nm`
C. `(0.01227)/(sqrt(V))nm`
D. `(12.27)/(sqrt(V))nm`

Explanation

To find the de Broglie wavelength \(\lambda\) of an electron accelerated through a potential \(V\) in volts, we can follow these steps:

### Step 1: Understand the relationship between kinetic energy and potential energy
When an electron is accelerated through a potential difference \(V\), it gains kinetic energy equal to the work done on it by the electric field. The kinetic energy \(K\) of the electron can be expressed as:
\[ K = eV \]
where \(e\) is the charge of the electron.

### Step 2: Relate kinetic energy to momentum
The kinetic energy can also be expressed in terms of momentum \(P\):
\[ K = \frac{P^2}{2m} \]
where \(m\) is the mass of the electron. Setting the two expressions for kinetic energy equal gives:
\[ eV = \frac{P^2}{2m} \]

### Step 3: Solve for momentum \(P\)
Rearranging the equation for momentum, we have:
\[ P^2 = 2m eV \]
Taking the square root of both sides gives:
\[ P = \sqrt{2m eV} \]

### Step 4: Use the de Broglie wavelength formula
The de Broglie wavelength \(\lambda\) is given by:
\[ \lambda = \frac{h}{P} \]
where \(h\) is Planck's constant. Substituting the expression for \(P\) into this equation, we get:
\[ \lambda = \frac{h}{\sqrt{2m eV}} \]

### Step 5: Substitute known values
Now, we can substitute the known values for \(h\), \(m\), and \(e\):
- \(h = 6.63 \times 10^{-34} \, \text{Js}\)
- \(m = 9.1 \times 10^{-31} \, \text{kg}\)
- \(e = 1.6 \times 10^{-19} \, \text{C}\)

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

### Step 6: Simplify the expression
Calculating the constants:
\[ \lambda = \frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19}}} \cdot \frac{1}{\sqrt{V}} \]
After calculating the constants, we find:
\[ \lambda \approx \frac{1.227 \times 10^{-9}}{\sqrt{V}} \, \text{m} \]
To express this in nanometers (nm), we multiply by \(10^9\):
\[ \lambda \approx \frac{1.227}{\sqrt{V}} \, \text{nm} \]

### Final Answer
Thus, the de Broglie wavelength \(\lambda\) of an electron accelerated through a potential \(V\) in volts is:
\[ \lambda = \frac{1.227}{\sqrt{V}} \, \text{nm} \]

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