A particle has a de Broglie wavelength lambda | vedclass

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

Question

(B) The de Broglie wavelength is given by $\lambda = \frac{h}{mv}$.Given that the velocity $v$ becomes $2v$ and the kinetic energy $E = \frac{1}{2}mv^

Vedclass · Accepted Answer

$\frac{\lambda}{2}$

Vedclass · Answer

(B) The de Broglie wavelength is given by $\lambda = \frac{h}{mv}$.Given that the velocity $v$ becomes $2v$ and the kinetic energy $E = \frac{1}{2}mv^2$ becomes $2E$.Let the new wavelength be $\lambda'$.Using the relation $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$,we have $\lambda \propto \frac{1}{\sqrt{E}}$.If the energy $E$ is doubled to $2E$,the new wavelength $\lambda' = \frac{\lambda}{\sqrt{2}}$.However,the question specifies that both velocity and energy are doubled. Since $E = \frac{1}{2}mv^2$,if $v$ is doubled,$E$ must become $4E$. If the problem states $E$ is doubled independently,it implies a change in mass or a specific constraint. Assuming the standard de Broglie relation $\lambda = \frac{h}{mv}$:If $v' = 2v$ and $m$ is constant,then $\lambda' = \frac{h}{m(2v)} = \frac{\lambda}{2}$.Given the options provided,the correct physical interpretation for a particle with constant mass is $\lambda' = \frac{\lambda}{2}$.

$A$ particle has a de Broglie wavelength $\lambda$. If its velocity and energy are both doubled,what will be the new de Broglie wavelength?

Similar Questions

$A$ stationary mass $M$ splits into two fragments of masses $m_1$ and $m_2$. What is the ratio of their de Broglie wavelengths,${\lambda _1}/{\lambda _2}$?

$A$ proton of mass $m_p$ has the same energy as that of a photon of wavelength $\lambda$. If the proton is moving at non-relativistic speed,then the ratio of its de Broglie wavelength to the wavelength of the photon is:

If the momentum of an electron is changed by $\Delta P,$ then the de Broglie wavelength associated with it changes by $0.5\%.$ The initial momentum of the electron will be

The de-Broglie wavelength $(\lambda)$ of a particle is related to its kinetic energy $(E)$ as

The de Broglie wavelengths for an electron and a photon are $\lambda_{e}$ and $\lambda_{p}$ respectively. For the same kinetic energy $K$ of the electron and the photon,which of the following presents the correct relation between the de Broglie wavelengths of the two?

Vedclass Test Series

Exam Paper Generator

Online Exam Module