The de Broglie wavelength and kinetic energy of a particle is `2000 Å and 1 eV` respectively. If its kinetic energy becomes 1 MeV, then its de Broglie wavelength becomes

Options

A. `1Å`
B. `2Å`
C. `5Å`
D. `10Å`

Explanation

To solve the problem, we need to use the relationship between de Broglie wavelength (λ) and kinetic energy (K.E.) of a particle. The de Broglie wavelength is given by the formula:

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

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

\[ K.E. = \frac{p^2}{2m} \implies p = \sqrt{2m \cdot K.E.} \]

Substituting this expression for momentum into the de Broglie wavelength formula gives:

\[ \lambda = \frac{h}{\sqrt{2m \cdot K.E.}} \]

### Step 1: Write down the initial conditions
We are given:
- Initial de Broglie wavelength \( \lambda_1 = 2000 \, \text{Å} = 2000 \times 10^{-10} \, \text{m} \)
- Initial kinetic energy \( K.E_1 = 1 \, \text{eV} \)

### Step 2: Write down the final conditions
We need to find the new de Broglie wavelength \( \lambda_2 \) when the kinetic energy becomes:
- Final kinetic energy \( K.E_2 = 1 \, \text{MeV} = 1 \times 10^6 \, \text{eV} \)

### Step 3: Set up the ratio of wavelengths
Using the relationship derived earlier, we can set up the ratio of the wavelengths based on the kinetic energies:

\[ \frac{\lambda_1}{\lambda_2} = \sqrt{\frac{K.E_2}{K.E_1}} \]

### Step 4: Substitute the values
Substituting the known values into the equation:

\[ \frac{2000 \times 10^{-10}}{\lambda_2} = \sqrt{\frac{1 \times 10^6}{1}} \]

### Step 5: Calculate the right side
Calculating the right side:

\[ \sqrt{1 \times 10^6} = 1000 \]

### Step 6: Rearranging the equation
Now we can rearrange the equation to solve for \( \lambda_2 \):

\[ \lambda_2 = \frac{2000 \times 10^{-10}}{1000} \]

### Step 7: Simplifying
Simplifying gives:

\[ \lambda_2 = 2 \times 10^{-10} \, \text{m} \]

### Step 8: Convert to angstroms
Since \( 1 \, \text{Å} = 10^{-10} \, \text{m} \):

\[ \lambda_2 = 2 \, \text{Å} \]

### Final Answer
Thus, the new de Broglie wavelength when the kinetic energy becomes 1 MeV is:

\[ \lambda_2 = 2 \, \text{Å} \]

---