Question

de-Broglie wavelength associated with an electron accelerated through a potential difference $V$ is $\lambda$. When the accelerating potential is increased to $4V$, the de-Broglie wavelength

• A. reduces to half
• B. remains the same
• C. reduces to $(1/4)\text{th}$
• D. increases by $25\%$

Hint:

Because $\lambda$ depends inversely on $\sqrt{V}$, changing the voltage by any factor $k$ will always scale the wavelength by a factor of $\frac{1}{\sqrt{k}}$. Increasing potential by $4\times$ cuts the wavelength by $\sqrt{4} = 2$.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The problem asks how the de-Broglie wavelength ($\lambda$) of a matter wave associated with an electron changes when the electrical potential difference ($V$) used to accelerate it is scaled up by a factor of 4.

Step 2: Key Formula or Approach:
An electron of mass $m$ and charge $e$ accelerated through potential $V$ acquires a kinetic energy $K = eV$.
The momentum is related to kinetic energy by $p = \sqrt{2mK} = \sqrt{2meV}$.
The de-Broglie wavelength formula is:
$$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2meV}}$$ This indicates an inverse square-root proportionality relation:
$$\lambda \propto \frac{1}{\sqrt{V}}$$

Step 3: Detailed Explanation:
Let the initial conditions be potential $V_1 = V$ and wavelength $\lambda_1 = \lambda$.
Let the final conditions be potential $V_2 = 4V$ and wavelength $\lambda_2$.
Setting up a ratio based on our proportionality relation:
$$\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{V_1}{V_2}}$$ Substitute the values into the relation:
$$\frac{\lambda_2}{\lambda} = \sqrt{\frac{V}{4V}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$$ $$\lambda_2 = \frac{\lambda}{2}$$ This indicates that the wavelength decreases to half of its original value.

Step 4: Final Answer:
The de-Broglie wavelength reduces to half, matching option (A).