Question

Circumférence of Bohr orbit in Hydrogen is 3.322×10^-10m ,velocity of the electron is​

Answer 1

Solution :

Given :-

Circumference of first bohr orbit of hydrogen = 3.322 × 10$^{-10}$ m

To Find :-

Velocity of electron in first bohr orbit of hydrogen.

Concept :-

This question is completely based on concept of 'quantization'.

As per Bohr's quantization rule,

$\underline{\boxed{\bf{\pink{\large{mvr=\dfrac{nh}{2\pi}}}}}}$

Terms indication :-

• m denotes mass of electron
• v denotes velocity of electron
• r denotes radius of orbit
• n denotes principle quantum number of bohr orbit
• h denotes plank constant

Calculation :-

$\mapsto\bf\:mvr=\dfrac{nh}{2\pi}\\ \\ \mapsto\sf\:2\pi{r}=\dfrac{nh}{mv}\\ \\ \red{\sf\dag\:perimeter=2\pi{r}=3.322\times 10^{-10} m}\\ \\ \mapsto\sf\:3.322\times 10^{-10}=\dfrac{1\times 6.626\times 10^{-34}}{9.109\times 10^{-31}\times v}\\ \\ \mapsto\sf\:v=\dfrac{6.626\times 10^{-34}}{3.026\times 10^{-40}}\\ \\ \mapsto\underline{\boxed{\bf{\purple{v=2.18\times 10^6\:mps}}}}$

Additional information :-

▪ Mass of electron =9.109×10$^{-31}$kg

▪ Value of plank constant = 6.626×10$^{-34}$Js