If the kinetic energy of a free electron doub | vedclass

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

Question

(C) The de-Broglie wavelength $\lambda$ is related to the kinetic energy $K$ of an electron by the formula $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK

Vedclass · Accepted Answer

$\frac{1}{\sqrt{2}}$

Vedclass · Answer

(C) The de-Broglie wavelength $\lambda$ is related to the kinetic energy $K$ of an electron by the formula $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$,where $h$ is Planck's constant and $m$ is the mass of the electron.Let the initial kinetic energy be $K_1 = K$ and the initial wavelength be $\lambda_1 = \frac{h}{\sqrt{2mK}}$.If the kinetic energy doubles,the new kinetic energy is $K_2 = 2K$.The new wavelength is $\lambda_2 = \frac{h}{\sqrt{2m(2K)}} = \frac{h}{\sqrt{2} \cdot \sqrt{2mK}} = \frac{1}{\sqrt{2}} \lambda_1$.Therefore,the de-Broglie wavelength changes by a factor of $\frac{1}{\sqrt{2}}$.

If the kinetic energy of a free electron doubles,its de-Broglie wavelength changes by the factor

Similar Questions

If the velocity of an electron increases,what will be the change in its de Broglie wavelength?

State the de-Broglie hypothesis.

Orbits of a particle moving in a circle are such that the perimeter of the orbit equals an integer number of de-Broglie wavelengths of the particle. For a charged particle moving in a plane perpendicular to a magnetic field,the radius of the $n^{th}$ orbital will therefore be proportional to

Explain the particle-wave (dual) nature of matter.

If the kinetic energy of a free electron doubles,its de-Broglie wavelength $\lambda$ changes by a factor of:

Vedclass Test Series

Exam Paper Generator

Online Exam Module