Question 331: Dual Nature of Matter and Radiation

Question

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

Accepted Answer

The de Broglie wavelength ($$\lambda$$) of a particle is given by the formula:
$$\lambda = \frac{h}{\sqrt{2meV}}$$
where:
$$h$$ is the Planck's constant,
$$m$$ is the mass of the electron,
$$e$$ is the charge of the electron, and
$$V$$ is the accelerating potential.
When the electron is accelerated through a potential $$V$$, the wavelength is given by:
$$\lambda_1 = \frac{h}{\sqrt{2meV}}$$
And the given condition states that this wavelength is $$4\lambda$$.
When the accelerating potential is increased to $$4V$$, the new wavelength ($$\lambda_2$$) can be calculated as:
$$\lambda_2 = \frac{h}{\sqrt{2me(4V)}}$$
Let's simplify $$\lambda_2$$:
$$\lambda_2 = \frac{h}{\sqrt{8meV}} = \frac{h}{2\sqrt{2meV}} = \frac{1}{2} \cdot \frac{h}{\sqrt{2meV}}$$
So, $$\lambda_2 = \frac{1}{2}\lambda_1 = \frac{1}{2} \cdot 4\lambda = 2\lambda$$.
Hence, the answer is Option D: $$2 \lambda$$.

Answer

$$\frac{\lambda}{4}$$

Answer

$$\frac{\lambda}{2}$$

Answer

$$\lambda$$

Answer

$$2 \lambda$$

Dual Nature of Matter and Radiation

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

Choose the correct answer: