What possibly can be the ratio of the de Broglie wavelength for two electrons each having zero initial weighing `200 g` and moving at a speed of `5m//hr` of the order of.

Options

A. `3 : 10`
B. `10 : 3`
C. `1 : 2`
D. `2 : 1`

Explanation

To find the ratio of the de Broglie wavelengths of two electrons accelerated through different potentials, we can follow these steps:

### Step 1: Understand the de Broglie wavelength formula
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
For an electron, the momentum \(p\) can also be expressed in terms of kinetic energy (KE):
\[ p = \sqrt{2m \cdot KE} \]
where \(m\) is the mass of the electron and \(KE\) is the kinetic energy.

### Step 3: Substitute momentum into the de Broglie wavelength formula
Substituting the expression for momentum into the de Broglie wavelength formula gives:
\[ \lambda = \frac{h}{\sqrt{2m \cdot KE}} \]

### Step 4: Calculate the ratio of wavelengths for two different kinetic energies
Let \(KE_1\) be the kinetic energy for the first electron (accelerated through 50 volts) and \(KE_2\) for the second electron (accelerated through 200 volts):
- \(KE_1 = 50 \, \text{eV}\)
- \(KE_2 = 200 \, \text{eV}\)

The ratio of the wavelengths is:
\[ \frac{\lambda_1}{\lambda_2} = \frac{h/\sqrt{2m \cdot KE_1}}{h/\sqrt{2m \cdot KE_2}} = \frac{\sqrt{KE_2}}{\sqrt{KE_1}} = \sqrt{\frac{KE_2}{KE_1}} \]

### Step 5: Substitute the values of kinetic energy
Now substituting the values:
\[ \frac{\lambda_1}{\lambda_2} = \sqrt{\frac{200 \, \text{eV}}{50 \, \text{eV}}} = \sqrt{4} = 2 \]

### Step 6: Conclusion
Thus, the ratio of the de Broglie wavelengths of the two electrons is:
\[ \frac{\lambda_1}{\lambda_2} = 2:1 \]

### Final Answer
The ratio of the de Broglie wavelengths for the two electrons is \(2:1\).

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