When a hydrogen atom is in its third excided state, what are the shortest and longest wavelengths of the photons it can emit?

Short Answer

Expert verified
The shortest wavelength of the photons that a hydrogen atom can emit when it's in its third excited state is 97.4 nm, and the longest wavelength is 1870 nm.

Step by step solution

01

Find the energy levels of a hydrogen atom

The formula for the energy levels of a hydrogen atom is given by: \[E_n = -\dfrac{13.6 \thinspace eV}{n^2}\] Where \(E_n\) is the energy of the nth level and n is the principal quantum number.
02

Determine the allowed energy transitions

When the hydrogen atom is in its third excited state (n=4), it can emit photons by transitioning to any of the lower energy levels (n=1, 2, or 3). We'll calculate the emitted photon's energy for each transition, as it corresponds to the energy difference between the initial and final energy levels.
03

Calculate the energy difference for each transition

For the transition from n=4 to n=1: \[ \Delta E_{41} = E_1 - E_4 = \left(-\dfrac{13.6}{1^2}\right) - \left(-\dfrac{13.6}{4^2}\right) = 13.6 - 0.85 = 12.75 \thinspace eV\] For the transition from n=4 to n=2: \[ \Delta E_{42} = E_2 - E_4 = \left(-\dfrac{13.6}{2^2}\right) - \left(-\dfrac{13.6}{4^2}\right) = 3.4 - 0.85 = 2.55 \thinspace eV\] For the transition from n=4 to n=3: \[ \Delta E_{43} = E_3 - E_4 = \left(-\dfrac{13.6}{3^2}\right) - \left(-\dfrac{13.6}{4^2}\right) = 1.51 - 0.85 = 0.66 \thinspace eV\]
04

Find the shortest and longest wavelengths using the Planck's equation

The Planck's equation is given by: \[E = \dfrac{hc}{\lambda}\] Where E is the energy of the emitted photon, h is the Planck's constant (6.626 x 10^(-34) Js), c is the speed of light (3 x 10^8 m/s), and λ is the wavelength. To find the shortest wavelength, we use the largest energy difference (12.75 eV): \[\lambda_{min} = \dfrac{hc}{E_{12.75}} = \dfrac{6.626 \times 10^{-34} \thinspace Js \times 3 \times 10^8 \thinspace m/s}{12.75 \thinspace eV \times 1.6 \times 10^{-19} \thinspace J/eV} = 9.74 \times 10^{-8} \thinspace m = 97.4 \thinspace nm\] To find the longest wavelength, we use the smallest energy difference (0.66 eV): \[\lambda_{max} = \dfrac{hc}{E_{0.66}} = \dfrac{6.626 \times 10^{-34} \thinspace Js \times 3 \times 10^8 \thinspace m/s}{0.66 \thinspace eV \times 1.6 \times 10^{-19} \thinspace J/eV} = 1.87 \times 10^{-6} \thinspace m = 1870 \thinspace nm\] So, the shortest wavelength of the photons that a hydrogen atom can emit when it's in its third excited state is 97.4 nm, and the longest wavelength is 1870 nm.

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