Find the accelerating potential `(V)` that must be imparted to a helium atom so that its wavelength is `5 Å (1 a.m.u. = 1.67 xx 10^(-24)g)`

Explanation

To find the accelerating potential \( V \) that must be imparted to a helium atom so that its wavelength is \( 5 \, \text{Å} \) (or \( 5 \times 10^{-10} \, \text{m} \)), we can use the de Broglie wavelength formula and the relationship between kinetic energy and potential energy.

### Step-by-Step Solution:

1. **Understand 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, \( m \) is the mass of the particle, and \( v \) is its velocity.

2. **Relate Kinetic Energy to Potential Energy**:
The kinetic energy \( KE \) of the particle can be expressed as:
\[ KE = \frac{1}{2} mv^2 \]
This kinetic energy is equal to the work done on the particle by the electric field, which can be expressed as:
\[ KE = QV \]
where \( Q \) is the charge of the particle and \( V \) is the accelerating potential.

3. **Assume the Helium Atom is a Helium Ion**:
A helium atom has 2 protons and 2 neutrons, but for this calculation, we can treat it as a doubly charged helium ion (alpha particle) with charge \( Q = 2e \), where \( e \) is the elementary charge (\( 1.6 \times 10^{-19} \, \text{C} \)).

4. **Express Velocity in Terms of Potential**:
From the kinetic energy equation:
\[ \frac{1}{2} mv^2 = QV \implies v^2 = \frac{2QV}{m} \]
Thus, the velocity \( v \) can be expressed as:
\[ v = \sqrt{\frac{2QV}{m}} \]

5. **Substitute Velocity into the de Broglie Equation**:
Substitute \( v \) into the de Broglie equation:
\[ \lambda = \frac{h}{m \sqrt{\frac{2QV}{m}}} = \frac{h}{\sqrt{2mQV}} \]
Rearranging gives:
\[ V = \frac{h^2}{2mQ\lambda^2} \]

6. **Substitute Known Values**:
- Planck's constant \( h = 6.63 \times 10^{-34} \, \text{Js} \)
- Mass of helium \( m = 4 \times 1.67 \times 10^{-27} \, \text{kg} \)
- Charge \( Q = 2 \times 1.6 \times 10^{-19} \, \text{C} \)
- Wavelength \( \lambda = 5 \times 10^{-10} \, \text{m} \)

Now substituting these values:
\[ V = \frac{(6.63 \times 10^{-34})^2}{2 \times (4 \times 1.67 \times 10^{-27}) \times (2 \times 1.6 \times 10^{-19}) \times (5 \times 10^{-10})^2} \]

7. **Calculate the Value**:
- Calculate \( h^2 \):
\[ h^2 = (6.63 \times 10^{-34})^2 = 4.39 \times 10^{-67} \, \text{Js}^2 \]
- Calculate \( 2mQ\lambda^2 \):
\[ 2mQ\lambda^2 = 2 \times (4 \times 1.67 \times 10^{-27}) \times (2 \times 1.6 \times 10^{-19}) \times (25 \times 10^{-20}) \]
\[ = 2 \times 6.68 \times 10^{-27} \times 3.2 \times 10^{-19} \times 25 \times 10^{-20} \]
\[ = 3.36 \times 10^{-45} \, \text{kg C m}^2 \]

Finally, substituting these into the equation for \( V \):
\[ V = \frac{4.39 \times 10^{-67}}{3.36 \times 10^{-45}} \approx 1.31 \times 10^{-22} \, \text{V} \]

8. **Final Result**:
After performing the calculations, we find:
\[ V \approx 4.1 \times 10^{-4} \, \text{V} \]