Through what potential difference must an electron pass to have a wavelength of `500A^(@)`.

Explanation

To find the potential difference through which an electron must pass to have a wavelength of \(500 \, \text{Å}\) (angstroms), we can use the de Broglie wavelength formula and the relationship between kinetic energy and potential difference. Here’s a step-by-step solution:

### Step 1: Understand the de Broglie wavelength formula
The de Broglie wavelength (\(\lambda\)) of a particle is given by the formula:
\[ \lambda = \frac{h}{mv} \]
where:
- \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\)),
- \(m\) is the mass of the electron (\(9.11 \times 10^{-31} \, \text{kg}\)),
- \(v\) is the velocity of the electron.

### Step 2: Relate kinetic energy to potential difference
The kinetic energy (KE) of the electron when it is accelerated through a potential difference \(V\) is given by:
\[ KE = eV \]
where:
- \(e\) is the charge of the electron (\(1.6 \times 10^{-19} \, \text{C}\)).

The kinetic energy can also be expressed in terms of mass and velocity:
\[ KE = \frac{1}{2} mv^2 \]

### Step 3: Set the two expressions for kinetic energy equal
Equating the two expressions for kinetic energy gives:
\[ eV = \frac{1}{2} mv^2 \]

### Step 4: Solve for velocity in terms of potential difference
From the equation above, we can express \(v\) as:
\[ v = \sqrt{\frac{2eV}{m}} \]

### Step 5: Substitute \(v\) into the de Broglie wavelength formula
Substituting this expression for \(v\) into the de Broglie wavelength formula gives:
\[ \lambda = \frac{h}{m \sqrt{\frac{2eV}{m}}} \]
This simplifies to:
\[ \lambda = \frac{h}{\sqrt{2meV}} \]

### Step 6: Rearrange to find the potential difference \(V\)
Rearranging the equation to solve for \(V\) gives:
\[ V = \frac{h^2}{2me\lambda^2} \]

### Step 7: Substitute known values
Now we can substitute the known values into the equation. First, convert the wavelength from angstroms to meters:
\[ \lambda = 500 \, \text{Å} = 500 \times 10^{-10} \, \text{m} = 5 \times 10^{-8} \, \text{m} \]
Now substituting the values:
- \(h = 6.626 \times 10^{-34} \, \text{Js}\)
- \(m = 9.11 \times 10^{-31} \, \text{kg}\)
- \(e = 1.6 \times 10^{-19} \, \text{C}\)
- \(\lambda = 5 \times 10^{-8} \, \text{m}\)

\[ V = \frac{(6.626 \times 10^{-34})^2}{2 \times (9.11 \times 10^{-31}) \times (1.6 \times 10^{-19}) \times (5 \times 10^{-8})^2} \]

### Step 8: Calculate the potential difference
Calculating the above expression:
\[ V \approx 6.03 \times 10^{-3} \, \text{V} \text{ or } 6.03 \, \text{mV} \]

### Final Answer
The potential difference through which the electron must pass to have a wavelength of \(500 \, \text{Å}\) is approximately \(6.03 \, \text{mV}\).

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