For the Bohr's first orbit of circumference 2 | 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 of an electron in an orbit is given by $mvr = \frac{nh}{2\pi}$.Rearranging this eq

Vedclass · Accepted Answer

$2\pi r$

Vedclass · Answer

(A) According to Bohr's quantization condition,the angular momentum of an electron in an orbit is given by $mvr = \frac{nh}{2\pi}$.Rearranging this equation,we get $2\pi r = n \left( \frac{h}{mv} \right)$.From the de-Broglie hypothesis,the wavelength $\lambda$ is given by $\lambda = \frac{h}{mv}$.Substituting this into the Bohr's condition,we get $2\pi r = n\lambda$.For the first orbit,$n = 1$.Therefore,$2\pi r = 1 \cdot \lambda$,which implies $\lambda = 2\pi r$.

For the Bohr's first orbit of circumference $2\pi r$,the de-Broglie wavelength of the revolving electron will be

Similar Questions

Let $A = \{1, 2, \{3, 4\}, 5\}$. Which of the following statements is incorrect and why?
$\varnothing \in A$

$CsBr$ has $b.c.c.$ structure with edge length $4.3 \ \mathring{A}$. The shortest inter-ionic distance between $Cs^{+}$ and $Br^{-}$ is .............. $\mathring{A}$.

Which of the following is an acidic salt?

Which of the following hormone levels causes the release of an ovum from the Graafian follicle (ovulation)?

The condition that the roots of $x^3-b x^2+c x-d=0$ are in geometric progression is

Vedclass Test Series

Exam Paper Generator

Online Exam Module