A proton and deuteron are accelerated by same potential difference.Find the ratio of their de-Broglie wavelengths.

Explanation

To find the ratio of the de Broglie wavelengths of a proton and a deuteron accelerated by the same potential difference, we can follow these steps:

### Step 1: Understand the de Broglie wavelength formula
The de Broglie wavelength (\( \lambda \)) of a particle is given by the formula:
\[ \lambda = \frac{h}{\sqrt{2mv}} \]
where:
- \( h \) is the Planck's constant,
- \( m \) is the mass of the particle,
- \( v \) is the velocity of the particle.

### Step 2: Determine the kinetic energy from the potential difference
When a charged particle is accelerated through a potential difference \( V \), it gains kinetic energy equal to the work done on it by the electric field:
\[ KE = qV \]
where \( q \) is the charge of the particle.

### Step 3: Relate kinetic energy to velocity
The kinetic energy can also be expressed in terms of mass and velocity:
\[ KE = \frac{1}{2} mv^2 \]
Setting the two expressions for kinetic energy equal gives:
\[ qV = \frac{1}{2} mv^2 \]
From this, we can solve for \( v \):
\[ v = \sqrt{\frac{2qV}{m}} \]

### Step 4: Substitute \( v \) into the de Broglie wavelength formula
Substituting the expression for \( v \) into the de Broglie wavelength formula:
\[ \lambda = \frac{h}{\sqrt{2m \cdot \sqrt{\frac{2qV}{m}}}} = \frac{h}{\sqrt{2 \cdot \frac{2qV}{m}}} = \frac{h}{\sqrt{\frac{4qV}{m}}} = \frac{h \sqrt{m}}{2\sqrt{qV}} \]

### Step 5: Calculate the de Broglie wavelengths for the proton and deuteron
- For the proton:
- Mass \( m_p = m \) (mass of proton)
- Charge \( q_p = e \) (charge of proton)

Thus:
\[ \lambda_p = \frac{h \sqrt{m_p}}{2\sqrt{eV}} \]

- For the deuteron:
- Mass \( m_d = 2m \) (mass of deuteron, which is approximately twice that of the proton)
- Charge \( q_d = e \) (charge of deuteron)

Thus:
\[ \lambda_d = \frac{h \sqrt{m_d}}{2\sqrt{eV}} = \frac{h \sqrt{2m}}{2\sqrt{eV}} \]

### Step 6: Find the ratio of the de Broglie wavelengths
Now, we can find the ratio of the de Broglie wavelengths:
\[ \frac{\lambda_p}{\lambda_d} = \frac{\frac{h \sqrt{m}}{2\sqrt{eV}}}{\frac{h \sqrt{2m}}{2\sqrt{eV}}} \]
This simplifies to:
\[ \frac{\lambda_p}{\lambda_d} = \frac{\sqrt{m}}{\sqrt{2m}} = \frac{1}{\sqrt{2}} \]

### Final Answer
The ratio of the de Broglie wavelengths of the proton to the deuteron is:
\[ \frac{\lambda_p}{\lambda_d} = \frac{1}{\sqrt{2}} \]