Question

speed of electron in a hydrogen atom in ground state is 2.2×10^6 m/s. then what is the speed of the electron in second excited state
​

Answer 1

⟾ V1 = 2.2 × 10^6 m/s

⟾ n1 = 1

⟾ n2 = 3 (second excited state)

⟾ V2 = ?

$\frac{v1}{v2} = \frac{n2}{n1}$

$\frac{2.2 \times {10}^{6} }{v2} = \frac{3}{1}$

$v2 = 7.33 \times {10}^{5}$

Answer 2

The speed of the electron for a hydrogen atom in second excited state is $1.1 \times 10^6} \ \text{m}/\text{s}$

Explanation:

Given Data

$\text{Speed of an electron in ground state (v}_A) = 2.2 \times 10^6 \text{m}/\text{s} \ \text{(n = 1)}$

Find the speed the electron in second excited state

$\text{Bohr's Formula} \Rightarrow \ \text{v}=\frac{2 \text{K} \text{e}^{2} \text{Z}}{\text{n h}}$

From the above formula, it is understood that velocity 'v' and orbital number 'n' are inversely proportional.

                                           $\text{v} \propto \frac{1}{\text{n}}$

$\Rightarrow \frac{\text{v}_\text{A}}{\text{v}_\text{B}} = \frac{\text{n}_\text{B}}{\text{n}_\text{A}}$

$\Rightarrow {\text{v}_\text{B}} = \frac{\text{n}_\text{A} \times\text{v}_\text{A}}{\text{n}_\text{B}}$

Substitute the respective values in above equation

$\Rightarrow {\text{v}_\text{B}} = \frac{1 \times 2.2 \times 10^6}{2}$

$\Rightarrow {\text{v}_\text{B}} = \frac{2.2 \times 10^6}{2}$

${\text{v}_\text{B}} = 1.1 \times 10^6} \ \text{m}/\text{s}$

Therefore the speed of the electron in second excited state is $1.1 \times 10^6} \ \text{m}/\text{s}$ for a hydrogen atom, when the speed of the electron in ground state is  $2.2 \times 10^6 \text{m}/\text{s}$ .

To Learn More

1) If the velocity of the electron in first orbit of H atom is 2.18×10^6m/s, what is its value in third orbit?

https://brainly.in/question/4338041

2) Calculate the velocity of electron in the first Bohr orbit of hydrogen atom. [Given:- Bohr radius = 0.529 Å, h = 6.626×10-34 Js, mass of the electron = 9.1×10-31 kg]

https://brainly.in/question/3829236