Determine the frequency and wavelength (in nm) of the light obsorded when the e- goes from n=1 to n=5

Frequency = 3.16 ×10¹⁴ Hz

λ = 0.95×10² nm

Explanation:

Energy associated with nth state is,

En =  -13.6/n²

For n = 1

E₁ = -13.6 / 1²

E₁ = -13.6/1

E₁ = -13.6 ev

Kinetic energy of electron = -E₁ = 13.6 ev

For n = 5

E₅ = -13.6 / 5²

E₅ = -13.6/25

E₅ = -0.544 ev

Kinetic energy of electron = -E₅ = 0.544 ev

Wavelength of radiation emitted:

E = hc/λ = E₅ - E₁

hc/λ = E₅ - E₁

by putting values,

6.63×10⁻³⁴Js × 3×10⁸m/s / λ = -0.544ev  - (-13.6 ev  )

6.63×10⁻³⁴ Js× 3×10⁸m/s / λ = 13.056 ev

λ = 6.63×10⁻³⁴ Js× 3×10⁸m/s  /13.056 ev

λ = 6.63×10⁻³⁴ Js× 3×10⁸m/s  /13.056 × 1.6×10⁻¹⁹ J

λ = 19.89 ×10⁻²⁶ Jm / 13.056 × 1.6×10⁻¹⁹ J

λ = 19.89 ×10⁻²⁶ Jm / 20.9 ×10⁻¹⁹ J

λ = 0.95×10⁻⁷ m

m to nm:

0.95×10⁻⁷ m ×10⁹nm/1 m

0.95×10² nm

Frequency:

Frequency = speed of electron / wavelength

by putting values,

Frequency = 3×10⁸m/s /0.95×10⁻⁷ m

Frequency = 3.16 ×10¹⁴ s⁻¹

s⁻¹ = Hz

Frequency = 3.16 ×10¹⁴ Hz

the ratio of the conjugate base to the acid required to attain the desired ph.

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