Question

The best answer for the wavelength of light emitted when an electron transitions from the fourth energy level to the first energy level in a hydrogen atom is:

a. 97.2 nm

b 972 nm

c. 9.72 nm

d. 9720 nm

e. .972 nm

Answer 1

E = -13.6 /n² eV; eV is the electron volt.

Given from fourth energy level to first energy level:

E= 13.6 (1/n₂² - 1/n₁²) = 13.6 {(1/16)-(1)} = 12.75 eV

Conversion of eV is 12.75eV* 1.6*10^-9 J/eV= 2.04*10^-8J

Wavelength lamda = h c / E = ( 6.626 x 10 ^{- 34} J s) (3.0 x 10^8m/_s)/ (2.04*10^-8J); h is planks constant, c velocity of light

    On solving, lamda = 9.75*10^-8 m = 97.5 nm;                             Conversion:1m= 10^-9 nm