Question

The energy that should be added to an electron to reduce its de-Broglie wavelength from $1nm$ to $0.5nm$ is

• A. Equal to the initial energy
• B. Two times the initial energy
• C. Three times the initial energy
• D. Four times the initial energy

Answer 1

The correct option is C

Three times the initial energy

Step 1: Given:

De-Broglie wavelengths ${\lambda }_1=1nm$ (initial) and ${\lambda }_2=0.5nm$ (final)

Step 2: Formula used:

De-Broglie wavelength, $\lambda \propto \frac{1}{\sqrt{E}}\Rightarrow E\propto \frac{1}{{\lambda }^2}$

E is the energy.

Step 3: Calculating the initial energy:

$E_1\propto \frac{1}{{{\lambda }_1}^2}$

Step 4: Calculating the final energy:

$E_2\propto \frac{1}{{{\lambda }_2}^2}$

Step 5: Dividing step 4 by step 3:

$\frac{E_2}{E_1}={\left(\frac{{\lambda }_1}{{\lambda }_2}\right)}^2={\left(\frac{1}{0.5}\right)}^2$

$\Rightarrow E_2=4E_1$

Step 6 Calculating the added energy:

$∆E=E_2-E_1\Rightarrow 4E_1-E_1=3E_1$

As a result, the energy required to reduce an electron's de-Broglie wavelength from $1nm$ to $0.5nm$ is three times the initial energy.

Thus, option C is the correct answer.