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Nimfa-mama [501]
3 years ago
13

Calculate the wavelength of light emitted when each of the following transitions occur in the hydrogen atom. What type of electr

omagnetic radiation is emitted in each transition? n=3 --> n=2
Chemistry
1 answer:
Alecsey [184]3 years ago
5 0

Answer:

The wavelength of the emitted photon will be approximately 655 nm, which corresponds to the visible spectrum.

Explanation:

In order to answer this question, we need to recall Bohr's formula for the energy of each of the orbitals in the hydrogen atom:

E_{n} = -\frac{m_{e}e^{4}}{2(4\pi\epsilon_{0})^2\hbar^{2}}\frac{1}{n^2} = E_{1}\frac{1}{n^{2}}, where:

[tex]m_{e}[tex] = electron mass

e = electron charge

[tex]\epsilon_{0}[tex] = vacuum permittivity

[tex]\hbar[tex] = Planck's constant over 2pi

n = quantum number

[tex]E_{1}[tex] = hydrogen's ground state = -13.6 eV

Therefore, the energy of the emitted photon is given by the difference of the energy in the 3d orbital minus the energy in the 2nd orbital:

[tex]E_{3} - E_{2} = -13.6 eV(\frac{1}{3^{2}} - \frac{1}{2^{2}})=1.89 eV[tex]

Now, knowing the energy of the photon, we can calculate its wavelength using the equation:

[tex]E = \frac{hc}{\lambda}[tex], where:

E = Photon's energy

h = Planck's constant

c = speed of light in vacuum

[tex]\lambda[tex] = wavelength

Solving for [tex]\lambda[tex] and substituting the required values:

[tex]\lambda = \frac{hc}{E} = \frac{1.239 eV\mu m}{1.89 eV}=0.655\mu m = 655 nm[tex], which correspond to the visible spectrum (The visible spectrum includes wavelengths between 400 nm and 750 nm).

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