Question

An excited H atom with the electron in energy level 6 emits a photon of wavelength 1094 nm. What energy level does the electron drop to

Answer 1

3

Step-by-step explanation:

1/λ = R (1/n1^2 -1/n2^2)

Where;

λ= wavelength = 1094 nm

R = Rydberg constant = 1.0974 * 10^7 m-1

n1 = lower energy level =?

n2= higher energy level =6

1/1094 *10-9= 1.0974 * 10^7(1/n1^2 - 1/36)

(1094 *10-9)^-1/1.0974 * 10^7 =(1/n1^2 - 1/36)

0.083 = (1/n1^2 - 1/36)

0.083 + 1/36 =1/n1^2

0.111 =1/n1^2

n1^2 = 9

n1 =3