What will be the ratio of de - Broglie wavelengths of proton and `alpha` - particle of same energy ?

Explanation

To find the ratio of de Broglie wavelengths of a proton and an alpha particle of the same energy, we can follow these steps:

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

### Step 2: Relate momentum to kinetic energy
The linear momentum (\( p \)) can be expressed in terms of kinetic energy (\( KE \)):
\[ p = \sqrt{2 m KE} \]
where \( m \) is the mass of the particle.

### Step 3: Write the de Broglie wavelength for both particles
For the proton:
\[ \lambda_p = \frac{h}{p_p} = \frac{h}{\sqrt{2 m_p KE}} \]
For the alpha particle:
\[ \lambda_{\alpha} = \frac{h}{p_{\alpha}} = \frac{h}{\sqrt{2 m_{\alpha} KE}} \]

### Step 4: Set up the ratio of wavelengths
Now, we can find the ratio of the de Broglie wavelengths of the proton and the alpha particle:
\[ \frac{\lambda_p}{\lambda_{\alpha}} = \frac{\frac{h}{\sqrt{2 m_p KE}}}{\frac{h}{\sqrt{2 m_{\alpha} KE}}} \]

### Step 5: Simplify the ratio
The \( h \) and \( \sqrt{2 KE} \) terms cancel out:
\[ \frac{\lambda_p}{\lambda_{\alpha}} = \frac{\sqrt{m_{\alpha}}}{\sqrt{m_p}} \]

### Step 6: Substitute the masses
The mass of an alpha particle (\( m_{\alpha} \)) is approximately 4 times the mass of a proton (\( m_p \)):
\[ m_{\alpha} = 4 m_p \]
Thus, substituting this into the ratio gives:
\[ \frac{\lambda_p}{\lambda_{\alpha}} = \frac{\sqrt{4 m_p}}{\sqrt{m_p}} = \frac{2 \sqrt{m_p}}{\sqrt{m_p}} = 2 \]

### Final Result
The ratio of the de Broglie wavelengths of the proton to the alpha particle is:
\[ \frac{\lambda_p}{\lambda_{\alpha}} = 2 : 1 \]

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