contestada

g The energy associated with the lowest n = 1 state in a hydrogen atom is – 13.6 eV. The wavelength of the emission line corresponding to the transition from n = 4 to n = 3 is

Respuesta :

pstnonsonjoku

Answer:

10.9 * 10^-7 m

Explanation:

We know that the Rydberg constant (R) is – 13.6 eV = -2.18 * 10^-18 J

ΔE= -R(1/nf - 1/ni)

Where;

nf = final state

ni= initial state

ΔE=  -2.18 * 10^-18(1/3 - 1/4)

ΔE= -2.18 * 10^-18(1/12)

ΔE= -1.81 * 10^-19 J

From;

ΔE=hc/λ

h = plank's constant = 6.6 * 10^-34 Js

c = 3 * 10^8 ms-1

λ = ?

λ = hc/ΔE

λ = 6.6 * 10^-34 * 3 * 10^8/ 1.81 * 10^-19

λ = 10.9 * 10^-7 m

Otras preguntas