A particle A of mass ‘m’ and charge ‘q’ is accelerated by a potential difference of 50 V. Another particle B of mass ‘4 m’ and charge ‘q’ is accelerated by a potential difference of 2500 V. The ratio of de-Broglie wavelengths `(lambda_A)/(lambda_B)` is close to :

Options

A. 14.14
B. 0.07
C. 10
D. 4.47

Explanation

To find the ratio of de-Broglie wavelengths \((\lambda_A/\lambda_B)\) for the two particles A and B, we will follow these steps:

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

### Step 2: Express momentum in terms of mass and velocity
The momentum \(p\) can be expressed as:
\[ p = mv \]
where \(m\) is the mass of the particle and \(v\) is its velocity.

### Step 3: Relate velocity to 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:
\[ \frac{1}{2}mv^2 = qV \]
From this, we can solve for \(v\):
\[ v = \sqrt{\frac{2qV}{m}} \]

### Step 4: Substitute velocity into the momentum expression
Substituting \(v\) into the momentum expression gives:
\[ p = m \sqrt{\frac{2qV}{m}} = \sqrt{2mqV} \]

### Step 5: Substitute momentum into the de-Broglie wavelength formula
Now substituting \(p\) back into the de-Broglie wavelength formula:
\[ \lambda = \frac{h}{\sqrt{2mqV}} \]

### Step 6: Calculate \(\lambda_A\) and \(\lambda_B\)
For particle A:
- Mass \(m_A = m\)
- Charge \(q_A = q\)
- Potential difference \(V_A = 50 \, V\)

\[ \lambda_A = \frac{h}{\sqrt{2mq \cdot 50}} \]

For particle B:
- Mass \(m_B = 4m\)
- Charge \(q_B = q\)
- Potential difference \(V_B = 2500 \, V\)

\[ \lambda_B = \frac{h}{\sqrt{2 \cdot 4m \cdot q \cdot 2500}} = \frac{h}{\sqrt{8mq \cdot 2500}} \]

### Step 7: Find the ratio \(\frac{\lambda_A}{\lambda_B}\)
Now we can find the ratio:
\[ \frac{\lambda_A}{\lambda_B} = \frac{\frac{h}{\sqrt{2mq \cdot 50}}}{\frac{h}{\sqrt{8mq \cdot 2500}}} \]
This simplifies to:
\[ \frac{\lambda_A}{\lambda_B} = \frac{\sqrt{8mq \cdot 2500}}{\sqrt{2mq \cdot 50}} \]

### Step 8: Simplify the expression
\[ \frac{\lambda_A}{\lambda_B} = \frac{\sqrt{8 \cdot 2500}}{\sqrt{2 \cdot 50}} = \frac{\sqrt{20000}}{\sqrt{100}} = \frac{\sqrt{200}}{1} = \sqrt{200} = 10\sqrt{2} \]

### Final Result
Thus, the ratio of the de-Broglie wavelengths is:
\[ \frac{\lambda_A}{\lambda_B} = 10\sqrt{2} \approx 14.14 \]