The potential energy of a particle of mass m is given by `U(x){{:(E_(0),,,0lexlt1),(0,,,xgt1):}`
`lambda_(1)` and `lambda_(2)` are the de-Broglie wavelength of the particle, when `0le x le1` and `x gt 1` respectively. If the total energy of particle is `2E_(0)`, then the ratio `(lambda_(1))/(lambda_(2))` will be

Options

A. 2
B. 1
C. `sqrt(2)`
D. `1//sqrt(2)`

Explanation

To solve the problem, we need to find the ratio of the de Broglie wavelengths \( \lambda_1 \) and \( \lambda_2 \) for a particle with a given potential energy function and total energy. Let's break down the steps:

### Step 1: Understand the potential energy function
The potential energy \( U(x) \) is given as:
- \( U(x) = E_0 \) for \( 0 \leq x < 1 \)
- \( U(x) = 0 \) for \( x > 1 \)

### Step 2: Determine the kinetic energy in both regions
The total energy \( E \) of the particle is given as \( 2E_0 \).

**For \( 0 \leq x < 1 \):**
- The kinetic energy \( K_1 \) can be calculated as:
\[ K_1 = E - U(x) = 2E_0 - E_0 = E_0 \]

**For \( x > 1 \):**
- The kinetic energy \( K_2 \) is:
\[ K_2 = E - U(x) = 2E_0 - 0 = 2E_0 \]

### Step 3: Use the de Broglie wavelength formula
The de Broglie wavelength \( \lambda \) is given by:
\[ \lambda = \frac{h}{p} \]
where \( p \) is the momentum of the particle. The momentum can be expressed in terms of kinetic energy \( K \) as:
\[ p = \sqrt{2mK} \]
Thus, the de Broglie wavelength can be rewritten as:
\[ \lambda = \frac{h}{\sqrt{2mK}} \]

### Step 4: Calculate \( \lambda_1 \) and \( \lambda_2 \)
**For \( 0 \leq x < 1 \):**
- Using \( K_1 = E_0 \):
\[ \lambda_1 = \frac{h}{\sqrt{2mE_0}} \]

**For \( x > 1 \):**
- Using \( K_2 = 2E_0 \):
\[ \lambda_2 = \frac{h}{\sqrt{2m(2E_0)}} = \frac{h}{\sqrt{4mE_0}} = \frac{h}{2\sqrt{mE_0}} \]

### Step 5: Find the ratio \( \frac{\lambda_1}{\lambda_2} \)
Now we can find the ratio:
\[ \frac{\lambda_1}{\lambda_2} = \frac{\frac{h}{\sqrt{2mE_0}}}{\frac{h}{2\sqrt{mE_0}}} \]
The \( h \) and \( \sqrt{mE_0} \) terms cancel out:
\[ \frac{\lambda_1}{\lambda_2} = \frac{2}{\sqrt{2}} = \sqrt{2} \]

### Final Answer
Thus, the ratio \( \frac{\lambda_1}{\lambda_2} \) is \( \sqrt{2} \).

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