An electron accelerated through a potential d | vedclass

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

Question

(B) The de-Broglie wavelength $\lambda$ of an electron accelerated through a potential $V$ is given by $\lambda = \frac{h}{\sqrt{2meV}}$.This implies

Vedclass · Accepted Answer

$\frac{9}{4}$

Vedclass · Answer

(B) The de-Broglie wavelength $\lambda$ of an electron accelerated through a potential $V$ is given by $\lambda = \frac{h}{\sqrt{2meV}}$.This implies $\lambda \propto \frac{1}{\sqrt{V}}$,or $V \propto \frac{1}{\lambda^2}$.For the first case,$V_1 \propto \frac{1}{\lambda^2}$.In the second case,the wavelength increases by $50 \%$,so the new wavelength $\lambda' = \lambda + 0.5\lambda = 1.5\lambda = \frac{3}{2}\lambda$.Thus,$V_2 \propto \frac{1}{(\frac{3}{2}\lambda)^2} = \frac{1}{\frac{9}{4}\lambda^2}$.Taking the ratio,$\frac{V_1}{V_2} = \frac{1/\lambda^2}{1/(\frac{9}{4}\lambda^2)} = \frac{9}{4}$.

An electron accelerated through a potential difference $V_1$ has a de-Broglie wavelength of $\lambda$. When the potential is changed to $V_2$,its de-Broglie wavelength increases by $50 \%$. The value of $\left(\frac{V_1}{V_2}\right)$ is equal to :

Similar Questions

If the de-Broglie wavelength at a temperature of $27^oC$ is $\lambda$,what will be the de-Broglie wavelength at $927^oC$?

The wavelength of a photon is equal to the wavelength of an electron moving with a kinetic energy of $1.5 \, eV$. What is the energy of the photon?

An electron of mass $m$ and charge $e$ initially at rest gets accelerated by a constant electric field $E$. The rate of change of de-Broglie wavelength of the electron at time $t$ is (Ignore relativistic effect) ($h=$ Planck's constant).

The de-Broglie wavelength of a particle accelerated with $150 \ V$ potential is $10^{-10} \ m$. If it is accelerated by $600 \ V$ potential difference,its wavelength will be: ............... $\mathring{A}$

$A$ particle is travelling $4$ times as fast as an electron. Assuming the ratio of the de-Broglie wavelength of the particle to that of the electron is $2:1$,the mass of the particle is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module