Question 349: 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: