Kinetic energy of a proton is equal to energy | vedclass

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

Question

(D) For a proton,the kinetic energy $E$ is related to its momentum $p$ by the equation $E = \frac{p^2}{2m}$,where $m$ is the mass of the proton.From t

Vedclass · Accepted Answer

$0.5$

Vedclass · Answer

(D) For a proton,the kinetic energy $E$ is related to its momentum $p$ by the equation $E = \frac{p^2}{2m}$,where $m$ is the mass of the proton.From this,the momentum is $p = \sqrt{2mE}$.The de-Broglie wavelength of the proton is $\lambda_1 = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$.For a photon,the energy $E$ is related to its wavelength $\lambda_2$ by the equation $E = \frac{hc}{\lambda_2}$,where $c$ is the speed of light.Thus,the wavelength of the photon is $\lambda_2 = \frac{hc}{E}$.Now,taking the ratio of the wavelengths:$\frac{\lambda_1}{\lambda_2} = \left( \frac{h}{\sqrt{2mE}} ight) 	imes \left( \frac{E}{hc} ight) = \frac{1}{c} \sqrt{\frac{E}{2m}}$.This simplifies to $\frac{\lambda_1}{\lambda_2} = \left( \frac{1}{c\sqrt{2m}} ight) E^{1/2}$.Comparing this with $\frac{\lambda_1}{\lambda_2} \propto E^n$,we find that $n = 1/2 = 0.5$.

Kinetic energy of a proton is equal to energy $E$ of a photon. Let $\lambda_1$ be the de-Broglie wavelength of the proton and $\lambda_2$ be the wavelength of the photon. If $\left(\frac{\lambda_1}{\lambda_2}\right) \propto E^{n}$,then the value of $n$ is:

Similar Questions

What is the energy of a neutron with a de Broglie wavelength of $1 \ \mathring{A}$?

The energy that should be added to an electron to reduce its de-Broglie wavelength from $\lambda$ to $\frac{\lambda}{2}$ is $n$ times the initial energy. The value of $n$ is

What is the de-Broglie wavelength of a bullet of mass $0.033$ kg travelling at the speed of $1$ km/s? $(h = 6.6 \times 10^{-34} \text{ Js})$

The de Broglie wavelength $\lambda$ associated with a proton increases by $25 \%$,if its momentum is decreased by $p_0$. The initial momentum was

$A$ proton and a photon both have the same energy of $E = 100 \,keV$. If the de Broglie wavelength of the proton is $\lambda_1$ and that of the photon is $\lambda_2$,then $\lambda_1/\lambda_2$ is proportional to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module