Question 328: Dual Nature of Matter and Radiation

Question

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

Accepted Answer

The de Broglie wavelength associated with an electron is given by the formula:
$\lambda = \frac{h}{p}$
where $h$ is Planck's constant and $p$ is the momentum of the electron.
For an electron accelerated through a potential difference $V$, the energy acquired by the electron is given by:
$eV = \frac{1}{2} mv^2$
where $e$ is the charge of the electron, $m$ is the mass of the electron, and $v$ is the velocity of the electron.
From the above equation, we can express the momentum $p$ in terms of the voltage $V$:
$p = \sqrt{2meV}$
Substituting this into the de Broglie wavelength equation, we get:
$\lambda = \frac{h}{\sqrt{2meV}}$
This shows that the de Broglie wavelength is inversely proportional to the square root of the accelerating potential:
$\lambda \propto \frac{1}{\sqrt{V}}$
Now, when the accelerating potential is increased to $4V$, the new wavelength $\lambda'$ will be:
$\lambda' = \frac{h}{\sqrt{2me(4V)}} = \frac{h}{2\sqrt{2meV}} = \frac{\lambda}{2}$
So, the de Broglie wavelength reduces to half. Therefore, the correct answer is:
Option A: reduces to half

Answer

reduces to half

Answer

remains the same

Answer

reduces to $(1 / 4)^{	ext {th }}$

Answer

increases by $25 \%$

Dual Nature of Matter and Radiation

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

Choose the correct answer: