If the kinetic energy of an electron is incre | 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 the relation: $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2m(KE)}}$.From this relation,we can see that $\lambda

Vedclass · Accepted Answer

half

Vedclass · Answer

(B) The de-Broglie wavelength is given by the relation: $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2m(KE)}}$.From this relation,we can see that $\lambda \propto \frac{1}{\sqrt{KE}}$.Let the initial kinetic energy be $KE_1$ and the final kinetic energy be $KE_2 = 4 	imes KE_1$.Let the initial wavelength be $\lambda_1$ and the final wavelength be $\lambda_2$.Then,$\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{KE_1}{KE_2}} = \sqrt{\frac{KE_1}{4 	imes KE_1}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.Therefore,the new wavelength becomes half of the original wavelength.

If the kinetic energy of an electron is increased four times,the wavelength of the de-Broglie wave associated with it would become

Similar Questions

What is the de-Broglie wavelength of an electron,in a hydrogen atom,moving in an orbit having a maximum magnetic quantum number $m = +2$ in units of $\mathring{A}$?

Can the wave nature of ordinary objects be detected? Why?

The wavelength associated with the electron moving in the first orbit of hydrogen atom with velocity $2.2 \times 10^6 \ ms^{-1}$ (in $nm$) is $\left(m_e=9.0 \times 10^{-31} \ kg, h=6.6 \times 10^{-34} \ Js\right)$

If the kinetic energy of an electron is $2.8 \times 10^{-23} \ J$,then the de Broglie wavelength will be ....... $(m_e = 9.1 \times 10^{-31} \ kg)$.

The De Broglie wavelength $\lambda$ of an electron in the $2^{nd}$ Bohr orbit is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module