Question

The energy of an electron in an orbit is given by En= (-2.18 × 10^-18)/n^2. How much energy is required to remove an electron from n=2 orbit. What is the wavelength associated with it?

Answer 1

Answer:

It's wavelength is below Cosmic ray (0.01A°)

Explanation:

En=(-2.18×10^-18)/4 (n=2) =-54.5/10^20 which is less than 0.01A°

Answer 2

Answer:

Given that the electron energy in hydrogen atom

E = (-2.18 × $10^{-18}$ )/n J

Energy required for ionization from n = 2 is given by

∆E = E – E

= [0- [(2.18 × $10^{-18}$ )/ 4] J

= 0.545 ×$10^{-18}$  J

∆E = 5.45 × $10^{-19}$ J

Now, λ = he/∆E

= (6.626 × $10^{-34}$ ) (3 × $10^{8}$ )/ 5.45 × $10^{-19}$

= 3647 × $10^{-10}$ m

= 3647 Å