To what effective potential a proton beam be subjected to give its protons a wavelength of `1xx10^(-10)m`.

Explanation

To find the effective potential to which a proton beam must be subjected to give its protons a wavelength of \(1 \times 10^{-10} \, \text{m}\), we can follow these steps:

### Step 1: Use the de Broglie wavelength formula
The de Broglie wavelength (\(\lambda\)) 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 proton (\(1.67 \times 10^{-27} \, \text{kg}\)),
- \(v\) is the velocity of the proton.

### Step 2: Relate kinetic energy to voltage
The kinetic energy (KE) of the proton can also be expressed in terms of voltage (V):
\[ KE = \frac{1}{2} mv^2 = qV \]
where \(q\) is the charge of the proton (\(1.6 \times 10^{-19} \, \text{C}\)).

### Step 3: Rearrange the equations
From the de Broglie wavelength formula, we can express \(v\) in terms of \(\lambda\):
\[ v = \frac{h}{m\lambda} \]
Substituting this expression for \(v\) into the kinetic energy equation gives:
\[ KE = \frac{1}{2} m \left(\frac{h}{m\lambda}\right)^2 \]
This simplifies to:
\[ KE = \frac{h^2}{2m\lambda^2} \]

### Step 4: Substitute into the voltage equation
Now we can equate the kinetic energy to the charge times voltage:
\[ qV = \frac{h^2}{2m\lambda^2} \]
Rearranging for \(V\) gives:
\[ V = \frac{h^2}{2mq\lambda^2} \]

### Step 5: Substitute known values
Now we can substitute the known values into the equation:
- \(h = 6.626 \times 10^{-34} \, \text{Js}\)
- \(m = 1.67 \times 10^{-27} \, \text{kg}\)
- \(q = 1.6 \times 10^{-19} \, \text{C}\)
- \(\lambda = 1 \times 10^{-10} \, \text{m}\)

Substituting these values into the equation for \(V\):
\[ V = \frac{(6.626 \times 10^{-34})^2}{2 \times (1.67 \times 10^{-27}) \times (1.6 \times 10^{-19}) \times (1 \times 10^{-10})^2} \]

### Step 6: Calculate the voltage
Calculating the numerator:
\[ (6.626 \times 10^{-34})^2 = 4.39 \times 10^{-67} \, \text{J}^2\text{s}^2 \]
Calculating the denominator:
\[ 2 \times (1.67 \times 10^{-27}) \times (1.6 \times 10^{-19}) \times (1 \times 10^{-10})^2 = 5.344 \times 10^{-56} \, \text{kg} \cdot \text{C} \cdot \text{m}^2 \]
Now substituting these into the voltage equation:
\[ V = \frac{4.39 \times 10^{-67}}{5.344 \times 10^{-56}} \approx 0.0823 \, \text{V} \]

### Final Answer
The effective potential to which the proton beam must be subjected is approximately \(0.0823 \, \text{V}\).

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