The energy that should be added to an electron to reduce its de - Broglie wavelength from one `nm to 0.5 nm` is

Options

A. four times the initial energy
B. equal to the initial energy
C. twice the initial energy
D. thrice the initial energy

Explanation

To solve the problem of how much energy should be added to an electron to reduce its de Broglie wavelength from 1 nm to 0.5 nm, we can follow these steps:

### Step 1: Understand the de Broglie Wavelength Formula
The de Broglie wavelength (λ) of a particle is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the particle.

### Step 2: Relate Momentum to Energy
The momentum \( p \) of a particle can be expressed in terms of its kinetic energy \( E \):
\[ p = \sqrt{2mE} \]
where \( m \) is the mass of the particle.

### Step 3: Substitute Momentum in the de Broglie Equation
Substituting the expression for momentum into the de Broglie wavelength formula gives:
\[ \lambda = \frac{h}{\sqrt{2mE}} \]

### Step 4: Set Up the Ratio of Wavelengths
We have two wavelengths:
- Initial wavelength \( \lambda_1 = 1 \, \text{nm} \)
- Final wavelength \( \lambda_2 = 0.5 \, \text{nm} \)

Using the ratio of the wavelengths:
\[ \frac{\lambda_2}{\lambda_1} = \frac{h/\sqrt{2mE_2}}{h/\sqrt{2mE_1}} = \frac{\sqrt{E_1}}{\sqrt{E_2}} \]

### Step 5: Simplify the Ratio
This simplifies to:
\[ \frac{\lambda_2}{\lambda_1} = \frac{\sqrt{E_1}}{\sqrt{E_2}} \]
Squaring both sides gives:
\[ \left(\frac{\lambda_2}{\lambda_1}\right)^2 = \frac{E_1}{E_2} \]

### Step 6: Substitute the Values
Substituting the values of the wavelengths:
\[ \left(\frac{0.5 \, \text{nm}}{1 \, \text{nm}}\right)^2 = \frac{E_1}{E_2} \]
This simplifies to:
\[ \frac{1}{4} = \frac{E_1}{E_2} \]

### Step 7: Express \( E_2 \) in Terms of \( E_1 \)
From the above equation, we can express \( E_2 \):
\[ E_2 = 4E_1 \]

### Step 8: Calculate the Energy Required
The energy that should be added to the electron, \( \Delta E \), is given by:
\[ \Delta E = E_2 - E_1 = 4E_1 - E_1 = 3E_1 \]

### Conclusion
The energy that should be added to the electron to reduce its de Broglie wavelength from 1 nm to 0.5 nm is \( 3E_1 \).