The accelerating potential that must be imparted to a proton beam to given it an effective wavelength of 0.05 nm is approximately

Options

A. 0.325 V
B. 5.205 V
C. 52.05 V
D. 3.25 V

Answer

To determine the accelerating potential that must be imparted to a proton beam to give it an effective wavelength of 0.05 nm, 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 the Planck 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: Rearrange the formula to find velocity
Rearranging the formula to solve for \(v\):

\[ v = \frac{h}{m\lambda} \]

### Step 3: Substitute the values
Substituting the known values into the equation:

\[ \lambda = 0.05 \, \text{nm} = 0.05 \times 10^{-9} \, \text{m} \]

Now substituting \(h\), \(m\), and \(\lambda\):

\[ v = \frac{6.626 \times 10^{-34}}{1.67 \times 10^{-27} \times 0.05 \times 10^{-9}} \]

### Step 4: Calculate the velocity
Calculating the above expression:

\[ v = \frac{6.626 \times 10^{-34}}{8.35 \times 10^{-37}} \approx 7.94 \times 10^{3} \, \text{m/s} \]

### Step 5: Relate kinetic energy to accelerating potential
The kinetic energy (KE) acquired by the proton when accelerated through a potential \(V\) is given by:

\[ KE = eV \]

where \(e\) is the charge of the proton (\(1.602 \times 10^{-19} \, \text{C}\)).

The kinetic energy can also be expressed as:

\[ KE = \frac{1}{2} mv^2 \]

### Step 6: Set the two expressions for kinetic energy equal
Setting the two expressions for kinetic energy equal gives us:

\[ eV = \frac{1}{2} mv^2 \]

### Step 7: Solve for the accelerating potential \(V\)
Rearranging to solve for \(V\):

\[ V = \frac{mv^2}{2e} \]

### Step 8: Substitute the values
Substituting \(m\), \(v\), and \(e\):

\[ V = \frac{(1.67 \times 10^{-27}) \times (7.94 \times 10^{3})^2}{2 \times (1.602 \times 10^{-19})} \]

### Step 9: Calculate the accelerating potential
Calculating the above expression:

\[ V = \frac{(1.67 \times 10^{-27}) \times (63.04 \times 10^{6})}{3.204 \times 10^{-19}} \approx 0.32859 \, \text{V} \]

### Step 10: Round to the appropriate significant figures
The final answer, rounded to three significant figures, is approximately:

\[ V \approx 0.325 \, \text{V} \]

### Conclusion
The accelerating potential that must be imparted to a proton beam to give it an effective wavelength of 0.05 nm is approximately **0.325 V**.

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