Question

kinetic energy of an electron in second Bohr orbit of Li^+2 ion
10.8ev
13.6ev
17ev
30.6ev​

Answer 1

Answer:

I think this may help you to understand

Answer 2

The value of kinetic energy is 5.4 x 10^-19 joules.

Explanation:

We are given that:

• Electron is present in = Second orbit i.e. n =2
• To Find: The kinetic energy of Li2"+ K.E = ?

Solution:

The formula to find kinetic energy = K.E = 1/2 mv^2

For velocity:

v =  2 π ke ^2 / nh

V = 2 x 3.14 x( 9 x 10^9 ) x ( 1.6 x 10x10^-19)^2 / 2 x 6.63 x 10^-34 )

v = 1.09 x 10^-6 m /s

K. E = 1 / 2 mv^2

K.E = 1 / 2 x ( 9.11 x 10^-31 Kg) x ( 1.09 x 10^-6 m /s)^2

K.E = 5.4 x 10^-19 J