The potential energy of particle of mass m varies as
`U(x)={(E_(0)"for"0lexle1),(0" for " gt 1):}`
The de Broglie wavelength of the particle in the range `0lexle1 " is " lamda_(1)` and that in the range `xgt1" is "lamda_(2)`.
If the total of the particle is `2E_(0)," find "lamda_(1)//lamda_(2)`.

Options

A. `sqrt(2)`
B. `sqrt(3)`
C. `sqrt((1)/(2))`
D. `sqrt((2)/(3))`

Explanation

To solve the problem, we need to find the ratio of the de Broglie wavelengths \( \frac{\lambda_1}{\lambda_2} \) for a particle of mass \( m \) with a given potential energy function. The potential energy \( U(x) \) is defined as follows:

\[ U(x) = \begin{cases} E_0 & \text{for } 0 \leq x \leq 1 \\ 0 & \text{for } x > 1 \end{cases} \]

The total energy \( E \) of the particle is given as \( 2E_0 \).

### Step 1: Determine the Kinetic Energy in Each Region

1. **For the region \( 0 \leq x \leq 1 \)**:
- The potential energy \( U = E_0 \).
- The total energy \( E = K + U \) gives us:
\[ 2E_0 = K + E_0 \implies K = 2E_0 - E_0 = E_0 \]

2. **For the region \( x > 1 \)**:
- The potential energy \( U = 0 \).
- The total energy \( E = K + U \) gives us:
\[ 2E_0 = K + 0 \implies K = 2E_0 \]

### Step 2: Calculate the de Broglie Wavelengths

The de Broglie wavelength \( \lambda \) is given by the formula:

\[ \lambda = \frac{h}{p} \]

where \( p \) is the momentum of the particle. The momentum can also be expressed in terms of kinetic energy \( K \):

\[ p = \sqrt{2mK} \]

Thus, the de Broglie wavelength can be expressed as:

\[ \lambda = \frac{h}{\sqrt{2mK}} \]

1. **For \( \lambda_1 \) in the region \( 0 \leq x \leq 1 \)**:
- Here, \( K = E_0 \):
\[ \lambda_1 = \frac{h}{\sqrt{2mE_0}} \]

2. **For \( \lambda_2 \) in the region \( x > 1 \)**:
- Here, \( K = 2E_0 \):
\[ \lambda_2 = \frac{h}{\sqrt{2m(2E_0)}} = \frac{h}{\sqrt{4mE_0}} = \frac{h}{2\sqrt{mE_0}} \]

### Step 3: Find the Ratio \( \frac{\lambda_1}{\lambda_2} \)

Now we can find the ratio of the two wavelengths:

\[ \frac{\lambda_1}{\lambda_2} = \frac{\frac{h}{\sqrt{2mE_0}}}{\frac{h}{2\sqrt{mE_0}}} \]

This simplifies to:

\[ \frac{\lambda_1}{\lambda_2} = \frac{h}{\sqrt{2mE_0}} \cdot \frac{2\sqrt{mE_0}}{h} = \frac{2}{\sqrt{2}} = \sqrt{2} \]

### Final Answer

Thus, the ratio of the de Broglie wavelengths is:

\[ \frac{\lambda_1}{\lambda_2} = \sqrt{2} \]

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