The de-Broglie wavelength of an electron in t | vedclass

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

Question

(A) According to Bohr's quantization condition,the angular momentum $J_n$ of an electron in the $n^{th}$ orbit is given by $J_n = \frac{n h}{2 \pi}$.T

Vedclass · Accepted Answer

$J_n \propto \lambda_n$

Vedclass · Answer

(A) According to Bohr's quantization condition,the angular momentum $J_n$ of an electron in the $n^{th}$ orbit is given by $J_n = \frac{n h}{2 \pi}$.This implies $J_n \propto n$.According to the de-Broglie hypothesis,the wavelength $\lambda_n$ is given by $\lambda_n = \frac{h}{p} = \frac{h}{mv_n}$.For the $n^{th}$ orbit,the circumference is $2 \pi r_n = n \lambda_n$.Since $r_n \propto n^2$ and $v_n \propto \frac{1}{n}$,the angular momentum $J_n = m v_n r_n$ is proportional to $n$.From the relation $2 \pi r_n = n \lambda_n$,we have $\lambda_n = \frac{2 \pi r_n}{n}$.Substituting $r_n \propto n^2$,we get $\lambda_n \propto \frac{n^2}{n} = n$.Since $J_n \propto n$ and $\lambda_n \propto n$,it follows that $J_n \propto \lambda_n$.

The de-Broglie wavelength of an electron in the $n^{th}$ Bohr orbit is $\lambda_n$ and the angular momentum is $J_n$,then:

Similar Questions

If an electron is revolving in its Bohr orbit having Bohr radius of $0.529 Å$,then the radius of the third orbit is

The frequency of the light emitted when an electron transitions from the $n=4$ to $n=2$ level in a hydrogen atom is $\frac{3}{7}$ times the frequency of a transition in a $Li^{2+}$ ion. Which transition in the $Li^{2+}$ ion corresponds to this?

In the Franck-Hertz experiment, the first dip in the current-voltage graph for hydrogen is observed at $10.2 \, V$. The wavelength of light emitted by the hydrogen atom when excited to the first excitation level is $\qquad$ $nm$.
(Given $hc = 1245 \, eV \cdot nm$, $e = 1.6 \times 10^{-19} \, C$).

Vedclass Test Series

Exam Paper Generator

Online Exam Module