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 /n^{2} eV; eV is the electron volt.

Given from fourth energy level to first energy level:

E= 13.6 (1/n_{2}^{2} -
1/n_{1}^{2}) = 13.6 {(1/16)-(1)} = 12.75 eV

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

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

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

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