Question

The electron in an hydrogen atom makes transition from K- shell to M- shell. The ratio of angular velocity in initial and final state is

Answer 1

ratio of angular velocity in initial and final state is 27 : 1 1.

It has given that the electron in an hydrogen atom makes transition from K - shell to M - shell.

we have to find the ratio of angular velocity in initial and final state.

using Bohr's atomic model,

r_n ∝ n²/Z ........(1)

v_n ∝ Z/n ............(2)

where r_n is the radius of nth orbit and Z is atomic number and v_n is the velocity of electron in nth orbit.

now angular velocity, ω_n = v_n/r_n ∝(Z/n)/(n²/Z) = Z²/n³

⇒ ω_n ∝ Z²/n³

here for hydrogen atom, Z = 1

so, angular velocity is directly proportional to 1/n³.

for K - shell, n = 1 , ω₁ = 1/1³ = 1

for M - shell, n = 3, ω₃ = 1/3³ = 1/27

therefore ratio of angular velocity in initial and final state is 27 : 1.

Answer 2

Given that,

The electron in an hydrogen atom makes transition from K- shell to M- shell.

We know that,

The transition from K shell to M shell

$K =1$

$M=3$

We need to calculate the angular velocity

Using Bohr's formula

$mvr=\dfrac{h}{2\pi}$

$v=\dfrac{h}{2\pi m\times r}$

We know that,

$r\propto n^2$

Put the value in to the formula

$v=\dfrac{h}{2\pi\times m\times n^2}$

$v\propto\dfrac{1}{n}$

We know that,

The angular velocity is

$\omega=\dfrac{v}{r}$

Put the value into the formula

$\omega=\dfrac{1}{n^3}$

We need to calculate the ratio of angular velocity in initial and final state

Using formula of velocity

$\dfrac{\omega_{i}}{\omega_{f}}=\dfrac{\dfrac{1}{n_{i}^3}}{\dfrac{1}{n_{f}^3}}$

$\dfrac{\omega_{i}}{\omega_{f}}=\dfrac{n_{f}^3}{n_{i}^3}$

Put the value into the formula

$\dfrac{v_{i}}{v_{f}}=\dfrac{3^3}{1^3}$

$\dfrac{v_{i}}{v_{f}}=\dfrac{27}{1}$

Hence, The ratio of angular velocity in initial and final state is 27:1.