Question

the ratio of radius of two different orbits in a h like atom is 4:9 the the the ratio of the frequency of the revolution of electron in these orbits are

Answer 1

Answer:

1) The ration between frequency and radius for an electron in an orbit is

ν=$\frac{v}{2\pi T }$

2) Thus the ratio is

$\frac{v_{1} }{v_{2} }$=$\frac{\frac{v}{2\pi4R } }{\frac{v}{2\pi9R } }$

=$\frac{4}{9}$

=4:9

Answer 2

Given:

Ratio of two different orbits = 4:9

To Find:

Frequency of the revolution of electron in these orbits

Solution:

The relation between frequency and radius for an electron in an orbit is-

v' = v/2πr

v ∝1/r

v1v2 = r2/r1

v1/v2 = 1/(r1/r2) -- eq 1

r1/r2 = 4/9 -- eq 2

From equations 1 and 2

v1/v2 = 1/(4/9)

= 9/4

Answer: The ratio of frequency of revolution of electron is 9:4