The de Broglie wavelength of a charged partic | vedclass

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

Question

(D) The de Broglie wavelength $\lambda$ of a charged particle accelerated through a potential difference $V$ is given by the formula:$\lambda = \frac{

Vedclass · Accepted Answer

$\frac{10 \lambda}{11}$

Vedclass · Answer

(D) The de Broglie wavelength $\lambda$ of a charged particle accelerated through a potential difference $V$ is given by the formula:$\lambda = \frac{h}{\sqrt{2mqV}}$From this expression,we can see that $\lambda \propto \frac{1}{\sqrt{V}}$.Given that the potential difference is increased by $21 \%$,the new potential difference $V^{\prime}$ is:$V^{\prime} = V + 0.21V = 1.21V$Let the new de Broglie wavelength be $\lambda^{\prime}$. Then:$\frac{\lambda^{\prime}}{\lambda} = \sqrt{\frac{V}{V^{\prime}}} = \sqrt{\frac{V}{1.21V}} = \sqrt{\frac{1}{1.21}} = \frac{1}{1.1} = \frac{10}{11}$Therefore,$\lambda^{\prime} = \frac{10}{11} \lambda$.

The de Broglie wavelength of a charged particle accelerated through a potential difference $V$ is $\lambda$. If the potential difference is increased by $21 \%$,the de Broglie wavelength of the charged particle is

Similar Questions

$A$ particle $A$ of mass $m$ and initial velocity $v$ collides with a particle $B$ of mass $\frac{m}{2}$ which is at rest. The collision is head-on and elastic. The ratio of the de-Broglie wavelengths $\lambda_A$ and $\lambda_B$ after the collision is

The de-Broglie wavelength of an electron moving with a velocity of $1.5 \times 10^8 \ m/s$ is equal to that of a photon. What is the ratio of the kinetic energy of the electron to that of the photon? (Given: $c = 3 \times 10^8 \ m/s$)

The resolving power of an electron microscope is greater than the resolving power of an optical microscope because

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 $4V$,its wavelength will be

An electron is accelerated from rest through a potential difference of $V$ volt. If the de Broglie wavelength of the electron is $1.227 \times 10^{-2} \, nm$,the potential difference is $......V$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module