The potential energy of a particle varies as .
`U(x) = E_0 ` for ` 0 le x le 1`
`= 0` for `x gt 1 `
for `0 le x le 1` de- Broglie wavelength is `lambda_1` and for `xgt1` the de-Broglie wavelength is `lambda_2`. Total energy of the particle is `2E_0`. find `(lambda_1)/(lambda_2).`

Options

A. `sqrt((3)`
B. `sqrt((7)`
C. `sqrt((2)`
D. `sqrt((5)`

Answer

To solve the problem, we need to find the ratio of the de Broglie wavelengths \( \frac{\lambda_1}{\lambda_2} \) for a particle in two different potential energy regions. Let's break down the solution step by step.

### Step 1: Analyze the potential energy in the two regions

The potential energy \( U(x) \) is defined as:
- \( U(x) = E_0 \) for \( 0 \leq x \leq 1 \)
- \( U(x) = 0 \) for \( x > 1 \)

The total energy of the particle is given as \( E_{total} = 2E_0 \).

### Step 2: Calculate the kinetic energy in the first region \( (0 \leq x \leq 1) \)

In the region where \( 0 \leq x \leq 1 \):
- The potential energy \( U = E_0 \)
- The total energy \( E_{total} = 2E_0 \)

Using the relationship between total energy, potential energy, and kinetic energy:
\[ K.E. = E_{total} - U = 2E_0 - E_0 = E_0 \]

### Step 3: Calculate the de Broglie wavelength \( \lambda_1 \) in the first region

The de Broglie wavelength is given by the formula:
\[ \lambda = \frac{h}{\sqrt{2mK.E.}} \]

Substituting the kinetic energy we found:
\[ \lambda_1 = \frac{h}{\sqrt{2mE_0}} \]

### Step 4: Calculate the kinetic energy in the second region \( (x > 1) \)

In the region where \( x > 1 \):
- The potential energy \( U = 0 \)
- The total energy \( E_{total} = 2E_0 \)

Using the relationship again:
\[ K.E. = E_{total} - U = 2E_0 - 0 = 2E_0 \]

### Step 5: Calculate the de Broglie wavelength \( \lambda_2 \) in the second region

Using the kinetic energy we found:
\[ \lambda_2 = \frac{h}{\sqrt{2m(2E_0)}} = \frac{h}{\sqrt{4mE_0}} \]

### Step 6: Find the ratio \( \frac{\lambda_1}{\lambda_2} \)

Now we can find the ratio of the two wavelengths:
\[ \frac{\lambda_1}{\lambda_2} = \frac{\frac{h}{\sqrt{2mE_0}}}{\frac{h}{\sqrt{4mE_0}}} \]

Simplifying this expression:
\[ \frac{\lambda_1}{\lambda_2} = \frac{h \cdot \sqrt{4mE_0}}{h \cdot \sqrt{2mE_0}} = \frac{\sqrt{4}}{\sqrt{2}} = \sqrt{2} \]

### Final Result

Thus, the ratio of the de Broglie wavelengths is:
\[ \frac{\lambda_1}{\lambda_2} = \sqrt{2} \]