Find the number of waves formed by an electro | vedclass

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

Question

(A) According to Bohr's postulate,the angular momentum of an electron in a stationary orbit is quantized as $mvr = \frac{nh}{2\pi}$.From de Broglie's

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(A) According to Bohr's postulate,the angular momentum of an electron in a stationary orbit is quantized as $mvr = \frac{nh}{2\pi}$.From de Broglie's hypothesis,the wavelength of an electron is $\lambda = \frac{h}{mv}$.Substituting $mv = \frac{h}{\lambda}$ into the Bohr's postulate: $\frac{h}{\lambda} 	imes r = \frac{nh}{2\pi}$.This simplifies to $2\pi r = n\lambda$,where $2\pi r$ is the circumference of the orbit.Therefore,the number of waves formed by an electron in an orbit is equal to the principal quantum number $n$.For the $n = 3$ orbit,the number of waves is $3$.

Find the number of waves formed by an electron in the $n = 3$ orbit of a Bohr atom.

Similar Questions

If the radius of the first orbit of the $H$ atom is $a_0$, then the de Broglie wavelength of the electron in the third orbit will be ....... (in $\pi a_0$)

Calculate the de Broglie wavelength of a $CO_2$ molecule moving with a velocity of $440 \, m/s$. (Molar mass of $CO_2 = 44 \, g/mol$)

The de Broglie wavelength $(\lambda)$ associated with a photoelectron varies with the frequency $(v)$ of the incident radiation as,[$v_0$ is threshold frequency]:

The de-Broglie's wavelength of an electron in the $4^{th}$ orbit is . . . . . . . . $\pi a_0$. ($a_0 =$ Bohr's radius)

The de-Broglie equation is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module