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: Understand the change.
An electron is accelerated through a voltage $V$ and behaves like a wave of wavelength $\lambda$. The voltage is then raised to $4V$, and we want the new wavelength.
Step 2: Link voltage to kinetic energy.
Work done by the field equals the kinetic energy gained: $K = eV$. A higher voltage gives a faster electron.
Step 3: Link kinetic energy to momentum.
For a non-relativistic electron $K = \dfrac{p^2}{2m}$, so $p = \sqrt{2mK} = \sqrt{2meV}$.
Step 4: Write the de-Broglie relation.
$\lambda = \dfrac{h}{p} = \dfrac{h}{\sqrt{2meV}}$. The only variable here is $V$, so $\lambda \propto \dfrac{1}{\sqrt{V}}$.
Step 5: Apply the ratio.
$\dfrac{\lambda_2}{\lambda_1} = \sqrt{\dfrac{V_1}{V_2}} = \sqrt{\dfrac{V}{4V}} = \sqrt{\dfrac{1}{4}} = \dfrac{1}{2}$.
Step 6: State the result.
Hence $\lambda_2 = \dfrac{\lambda}{2}$: the wavelength is halved when the voltage is quadrupled, which is option (1). The inverse-square-root law explains why a four-fold voltage gives only a two-fold change.
\[ \boxed{\lambda_2 = \dfrac{\lambda}{2}\ \text{(reduces to half)}} \]