We study the atomic emission spectra of hydrogen.

1) What is the energy of an electron in the third orbital (n =3)?

2) Calculate the energy of the n=3 to n=1 atomic transition.

3) What is the energy of a photon emitted at λ = 656 nm?

4) The retina of the human eye can detect light with energy ≥ 4.0 x 10^-17 J.
What is the minimum number of photons at λ = 656 nm that must be emitted for a person to see the red emission line?

Data: Rydberg constant = -2.1799 aJ, h = 6.626 x 10^-34 J.s, c = 2.998 x 10⁸ m.s^-1

Question

We study the atomic emission spectra of hydrogen.

1) What is the energy of an electron in the third orbital (n =3)?

2) Calculate the energy of the n=3 to n=1 atomic transition.

3) What is the energy of a photon emitted at λ = 656 nm?

4) The retina of the human eye can detect light with energy ≥ 4.0 x 10^-17 J.
What is the minimum number of photons at λ = 656 nm that must be emitted for a person to see the red emission line?

Data: Rydberg constant = -2.1799 aJ, h = 6.626 x 10^-34 J.s, c = 2.998 x 10⁸ m.s^-1

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Accepted Answer

1) E = $\frac{- 2.1799 (aJ)}{n^{2}}$ = - 2.4221 x 10^-1aJ = -2.4221 x 10^-19 J

2) E_photon = 2.1799 ( $\frac{1}{{n_{f}}^{2}}$ – $\frac{1}{{n_{i}}^{2}}$ ) = 2.1799 (1 – $\frac{1}{9}$ ) = 1.9377 aJ = 1.9377 x 10^-18 J

3) E_photon-656 = hν = $\frac{hc}{λ}$ = $\frac{(6.626 \times 10^{- 34}) \times (2.998 \times 10^{8})}{656 \times 10^{- 9}}$ = 3.028 x 10^-19 J

4) E_total ≥ 4.0 x 10^-17 J

and E_total = n x E_photon-656⇒n = E_total/ E_photon-656

n ≥ 4.0 x 10^-17 / 3.028 x 10^-19

n ≥ 132.1

The minimum number of photons is 133.

Final - Question 9 | General Chemistry 1

General Chemistry 1 - Final