Question

What is the additional energy that should be supplied to a moving electron to reduce its de Broglie wavelength from $1\ \text{nm}$ to $0.5\ \text{nm}$?

• A. Four times its initial energy
• B. Five times its initial energy
• C. Two times its initial energy
• D. Three times its initial energy

Hint:

Always read carefully to see if a question asks for the "final" amount or the "additional" amount. The final energy is $4E_1$, making (A) a very tempting trap, but the additional energy is only $3E_1$.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We need to find the amount of

extra kinetic energy required to halve the de Broglie wavelength of an electron.

Step 2: Key Formula or Approach:
The de Broglie wavelength ($\lambda$) is related to the kinetic energy ($E$) of a particle by the formula:
$$\lambda = \frac{h}{\sqrt{2mE}}$$ This shows an inverse square relationship: $E \propto \frac{1}{\lambda^2}$.

Step 3: Detailed Explanation:
Let the initial wavelength be $\lambda_1 = 1\ \text{nm}$ and the initial energy be $E_1$.
Let the final wavelength be $\lambda_2 = 0.5\ \text{nm}$ and the final energy be $E_2$.
Using the proportionality $E \propto \frac{1}{\lambda^2}$, we can set up a ratio:
$$\frac{E_2}{E_1} = \left(\frac{\lambda_1}{\lambda_2}\right)^2$$ Substitute the given wavelength values:
$$\frac{E_2}{E_1} = \left(\frac{1}{0.5}\right)^2 = (2)^2 = 4$$ This tells us the final total energy is four times the initial energy:
$$E_2 = 4E_1$$ The question specifically asks for the

additional energy supplied, which is the difference between final and initial energy:
$$\Delta E = E_2 - E_1 = 4E_1 - E_1 = 3E_1$$

Step 4: Final Answer: The additional energy supplied must be three times its initial energy, matching option (D).