A deuteron and an `alpha` - partical have same kinetic energy. Find the ratio of their de-Broglie wavelengths.

Explanation

To find the ratio of the de-Broglie wavelengths of a deuteron and an alpha particle, given that they have the same kinetic energy, we can follow these steps:

### Step 1: Write the formula for de-Broglie wavelength
The de-Broglie wavelength (λ) of a particle is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the particle.

### Step 2: Relate momentum to kinetic energy
The momentum \( p \) of a particle can be expressed in terms of its kinetic energy \( K \) as follows:
\[ p = \sqrt{2mK} \]
where \( m \) is the mass of the particle and \( K \) is its kinetic energy.

### Step 3: Write the de-Broglie wavelength for both particles
For the deuteron (mass \( m_d \)):
\[ \lambda_d = \frac{h}{\sqrt{2m_d K}} \]
For the alpha particle (mass \( m_\alpha \)):
\[ \lambda_\alpha = \frac{h}{\sqrt{2m_\alpha K}} \]

### Step 4: Find the ratio of the wavelengths
Now we can find the ratio of the de-Broglie wavelengths:
\[ \frac{\lambda_d}{\lambda_\alpha} = \frac{\frac{h}{\sqrt{2m_d K}}}{\frac{h}{\sqrt{2m_\alpha K}}} \]
This simplifies to:
\[ \frac{\lambda_d}{\lambda_\alpha} = \frac{\sqrt{2m_\alpha K}}{\sqrt{2m_d K}} = \frac{\sqrt{m_\alpha}}{\sqrt{m_d}} \]

### Step 5: Substitute the mass values
The mass of the alpha particle is approximately \( 4m_n \) (where \( m_n \) is the mass of a neutron), and the mass of the deuteron is approximately \( 2m_n \). Thus:
\[ m_\alpha = 4m_n \quad \text{and} \quad m_d = 2m_n \]
Substituting these values into the ratio gives:
\[ \frac{\lambda_d}{\lambda_\alpha} = \frac{\sqrt{4m_n}}{\sqrt{2m_n}} = \frac{2\sqrt{m_n}}{\sqrt{2m_n}} = \frac{2}{\sqrt{2}} = \sqrt{2} \]

### Final Result
Thus, the ratio of the de-Broglie wavelengths of the deuteron to the alpha particle is:
\[ \frac{\lambda_d}{\lambda_\alpha} = \sqrt{2} \]