If the potential difference used to accelerat | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The de Broglie wavelength $\lambda$ of an electron accelerated through a potential difference $V$ is given by the formula: $\lambda = \frac{h}{\sq

Vedclass · Accepted Answer

Wavelength is decreased to $1/\sqrt{2}$ times.

Vedclass · Answer

(D) The de Broglie wavelength $\lambda$ of an electron accelerated through a potential difference $V$ is given by the formula: $\lambda = \frac{h}{\sqrt{2meV}}$.From this relation,we can see that $\lambda \propto \frac{1}{\sqrt{V}}$.Let the initial potential difference be $V_1 = V$ and the initial wavelength be $\lambda_1$.When the potential difference is doubled,the new potential difference is $V_2 = 2V$.The new wavelength $\lambda_2$ is given by $\lambda_2 = \frac{h}{\sqrt{2me(2V)}} = \frac{1}{\sqrt{2}} 	imes \frac{h}{\sqrt{2meV}}$.Therefore,$\lambda_2 = \frac{1}{\sqrt{2}} \lambda_1$.Thus,the de-Broglie wavelength changes by a factor of $1/\sqrt{2}$.

If the potential difference used to accelerate electrons is doubled,by what factor does the de-Broglie wavelength associated with the electrons change?

Similar Questions

Explain how,according to Born's probability interpretation,a wave having a single (unique) wavelength is extended all over space.

An electron accelerated through a potential of $10000 \ V$ from rest has a de-Broglie wavelength $\lambda$. What should be the accelerating potential,so that the wavelength is doubled (in $V$)?

$A$ particle is dropped from a height $H$. The de Broglie wavelength of the particle depends on height as

An $\alpha$-particle moves in a circular path of radius $1 \ cm$ in a uniform magnetic field of $0.125 \ T$. The de Broglie wavelength associated with the $\alpha$-particle is

Vedclass Test Series

Exam Paper Generator

Online Exam Module