If the radius of the first Bohr orbit is r,th | vedclass

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

Question

(C) According to Bohr's second postulate,the angular momentum is given by $\frac{nh}{2 \pi} = mvr_n$.From this,the de-Broglie wavelength $\lambda_n$ i

Vedclass · Accepted Answer

$8 \pi r$

Vedclass · Answer

(C) According to Bohr's second postulate,the angular momentum is given by $\frac{nh}{2 \pi} = mvr_n$.From this,the de-Broglie wavelength $\lambda_n$ is $\lambda_n = \frac{h}{mv} = \frac{2 \pi r_n}{n}$.We know that the radius of the $n^{	ext{th}}$ Bohr orbit is given by $r_n = r_1 	imes n^2$,where $r_1 = r$.For the $4^{	ext{th}}$ orbit $(n = 4)$,the radius is $r_4 = r 	imes 4^2 = 16r$.Substituting these values into the wavelength formula:$\lambda_4 = \frac{2 \pi r_4}{4} = \frac{2 \pi 	imes (16r)}{4} = 8 \pi r$.

If the radius of the first Bohr orbit is $r$,then the de-Broglie wavelength of the electron in the $4^{\text{th}}$ orbit will be:

Similar Questions

Mention the value of the Rydberg constant with its unit.

$A$ diatomic molecule is made of two masses $m_1$ and $m_2$ separated by a distance $r$. Applying the Bohr's quantization rule for angular momentum,calculate its rotational kinetic energy. It is given by the formula:

$A$ diatomic molecule is made of two masses $m_1$ and $m_2$ which are separated by a distance $r$. If we calculate its rotational energy by applying Bohr's rule of angular momentum quantization,its energy will be given by: ($n$ is an integer)

An electron in the $n = 1$ orbit of a hydrogen atom is bound by $13.6\, eV$. The energy required to ionize it is........$ eV$.

The ionization energy of a $Li^{++}$ ion is .......

Vedclass Test Series

Exam Paper Generator

Online Exam Module