Which of the following transitions represent the emission of a photon with the largest energy?
1 answer:
Answer:
n=6 to n=3 (B)
Explanation:
Energy of an electron present in the orbit is directly proportional to
.Hence a transistion from one orbit to another orbit emits an energy proportional to the difference of their squares of the orbits. that is if an electron travels from orbit n1 to orbit n2 then it emits an energy corresponding to
.So in the above question the highest energy emission occurs when an electron moves from n=6 to n=3.(Highest difference of energy levels).
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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:
, 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).