Question

9. The first energy level for an element lies at 1.154 x 10-20 1 and the second energy level lies at 3.460 x 10-20 J. Determine the wavelength and frequency of the photon emitted when an electron drops from the second energy level to the first level. In which region of the electromagnetic spectrum will I find that photon.

Answer 1

Difference in energy = 3.460*10^-20 J - 1.154*10^-20 J

= 2.306*10^-20 J

1)

use:

E = h*c/lambda

2.306*10^-20J =(6.626*10^-34 J.s)*(3.0*10^8 m/s)/lambda

lambda = 8.62*10^-6 m

Answer: 8.62*10^-6 m

2)

Given:

E = 2.306*10^-20 J

use:

E = h*f

2.306*10^-20 J =(6.626*10^-34 J.s)*f

f = 3.48*10^13 Hz

Answer: 3.48*10^13 Hz

3)

It lies in ultraviolet region