The de-Broglie wavelength associated with the | vedclass

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

Question

(B) The de-Broglie wavelength $\lambda$ of an electron is given by $\lambda = \frac{h}{mv}$.According to Bohr's theory,the velocity $v$ of an electron

Vedclass · Accepted Answer

four times the de-Broglie wavelength of the electron in the ground state

Vedclass · Answer

(B) The de-Broglie wavelength $\lambda$ of an electron is given by $\lambda = \frac{h}{mv}$.According to Bohr's theory,the velocity $v$ of an electron in the $n^{th}$ orbit is inversely proportional to the principal quantum number $n$,i.e.,$v \propto \frac{1}{n}$.Substituting this into the wavelength formula,we get $\lambda = \frac{h}{m(k/n)} = \frac{h}{mk} \cdot n$,which implies $\lambda \propto n$.For the ground state $(n = 1)$,let the wavelength be $\lambda_1$.For the $n = 4$ level,the wavelength $\lambda_4$ is given by $\lambda_4 = 4 \lambda_1$.Therefore,the de-Broglie wavelength in the $n = 4$ level is four times the de-Broglie wavelength in the ground state.

The de-Broglie wavelength associated with the electron in the $n = 4$ level is

Similar Questions

It is found experimentally that $13.6 \; eV$ energy is required to separate a hydrogen atom into a proton and an electron. Compute the orbital radius and the velocity of the electron in a hydrogen atom.

In the Bohr's hydrogen atom model,the radius of the stationary orbit is directly proportional to ($n =$ principle quantum number)

The orbital frequency of an electron in the hydrogen atom is proportional to

The concept of stationary orbits was proposed by

Vedclass Test Series

Exam Paper Generator

Online Exam Module