Number of waves produced by an electron in on | vedclass

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

Question

(A) The number of waves produced by an electron in one complete revolution is given by the ratio of the circumference of the orbit to the de Broglie w

Vedclass · Accepted Answer

$n$

Vedclass · Answer

(A) The number of waves produced by an electron in one complete revolution is given by the ratio of the circumference of the orbit to the de Broglie wavelength of the electron.Number of waves $= \frac{	ext{Circumference}}{\lambda} = \frac{2 \pi r}{\lambda}$According to de Broglie's hypothesis,$\lambda = \frac{h}{mv}$.Substituting this into the expression,we get: Number of waves $= \frac{2 \pi r}{(h / mv)} = \frac{2 \pi mvr}{h}$.According to Bohr's postulate,the angular momentum $mvr = \frac{nh}{2 \pi}$.Therefore,substituting $mvr$ into the expression: Number of waves $= \frac{2 \pi}{h} 	imes \frac{nh}{2 \pi} = n$.Thus,the number of waves is equal to the principal quantum number $n$.

Number of waves produced by an electron in one complete revolution in $n^{th}$ orbit is

Similar Questions

$A$ $0.66 \ kg$ ball is moving with a speed of $100 \ m/s$. The associated wavelength will be $(h = 6.6 \times 10^{-34} \ J \ s)$.

If the kinetic energy of an electron is $18.2 \times 10^{-25} \ J$,its de Broglie wavelength in $nm$ is: (mass of electron $= 9.1 \times 10^{-31} \ kg$; $h = 6.626 \times 10^{-34} \ J \ s$)

If the radius of the first orbit of $H$ atom is $a_0$, what is the de-Broglie wavelength of an electron in the third orbit (in $\pi a_0$)?

If the velocity of a hydrogen molecule is $5 \times 10^4 \, cm \, sec^{-1}$,then its de-Broglie wavelength is ............. $\mathring{A}$.

The de-Broglie equation is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module