The ratio of de-Broglie wavelength associated with two electrons accelerated through 25V and 36V is

Options

A. `25//36`
B. `36//25`
C. `5//6`
D. `6//5`

Explanation

To find the ratio of the de-Broglie wavelengths associated with two electrons accelerated through 25V and 36V, we can follow these steps:

### Step 1: Understand the relationship between kinetic energy and potential difference
When an electron is accelerated through a potential difference (V), its kinetic energy (KE) can be expressed as:
\[ KE = eV \]
where \( e \) is the charge of the electron.

### Step 2: Calculate the kinetic energy for both electrons
For the first electron accelerated through 25V:
\[ KE_1 = e \times 25 \]

For the second electron accelerated through 36V:
\[ KE_2 = e \times 36 \]

### Step 3: Relate kinetic energy to momentum
The kinetic energy can also be expressed in terms of momentum (p):
\[ KE = \frac{p^2}{2m} \]
where \( m \) is the mass of the electron. Rearranging gives us:
\[ p = \sqrt{2m \cdot KE} \]

### Step 4: Calculate the momentum for both electrons
For the first electron:
\[ p_1 = \sqrt{2m \cdot (e \times 25)} = \sqrt{50me} \]

For the second electron:
\[ p_2 = \sqrt{2m \cdot (e \times 36)} = \sqrt{72me} \]

### Step 5: Use the de-Broglie wavelength formula
The de-Broglie wavelength (\( \lambda \)) is given by:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant.

### Step 6: Calculate the de-Broglie wavelengths for both electrons
For the first electron:
\[ \lambda_1 = \frac{h}{p_1} = \frac{h}{\sqrt{50me}} \]

For the second electron:
\[ \lambda_2 = \frac{h}{p_2} = \frac{h}{\sqrt{72me}} \]

### Step 7: Find the ratio of the de-Broglie wavelengths
Now, we can find the ratio \( \frac{\lambda_1}{\lambda_2} \):
\[ \frac{\lambda_1}{\lambda_2} = \frac{\frac{h}{\sqrt{50me}}}{\frac{h}{\sqrt{72me}}} = \frac{\sqrt{72me}}{\sqrt{50me}} = \frac{\sqrt{72}}{\sqrt{50}} = \frac{\sqrt{36 \cdot 2}}{\sqrt{25 \cdot 2}} = \frac{6}{5} \]

### Conclusion
Thus, the ratio of the de-Broglie wavelengths associated with the two electrons is:
\[ \frac{\lambda_1}{\lambda_2} = \frac{6}{5} \]