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: Read the question carefully.
We must find the EXTRA energy needed to change an electron's de Broglie wavelength from $1$ nm to $0.5$ nm. "Extra" means the difference, not the final energy.

Step 2: Recall the wavelength formula.
The de Broglie wavelength in terms of kinetic energy $E$ is \[ \lambda=\frac{h}{\sqrt{2mE}}. \]

Step 3: See how energy and wavelength connect.
Squaring and rearranging shows $E\propto\frac{1}{\lambda^2}$. So a smaller wavelength means a larger energy, and the link is an inverse square.

Step 4: Set up the ratio.
With $\lambda_1=1$ nm (energy $E_1$) and $\lambda_2=0.5$ nm (energy $E_2$): \[ \frac{E_2}{E_1}=\left(\frac{\lambda_1}{\lambda_2}\right)^2. \]

Step 5: Put numbers in.
\[ \frac{E_2}{E_1}=\left(\frac{1}{0.5}\right)^2=(2)^2=4, \] so $E_2=4E_1$. The final energy is four times the start.

Step 6: Find the additional energy.
Extra energy is final minus initial: \[ \Delta E=E_2-E_1=4E_1-E_1=3E_1. \]

Step 7: State the result.
The extra energy supplied is three times the initial energy, which is option (4).
\[ \boxed{\Delta E=3E_1} \]