The electron in the hydrogen atom jumps from excited state (n=3) to its ground state (n=1) and the photons thus emitted irradiate a photosensitive material. If the work function of the material is 5.1eV, the stopping potential is estimated to be
Question
The electron in the hydrogen atom jumps from excited state (n=3) to its ground state (n=1) and the photons thus emitted irradiate a photosensitive material. If the work function of the material is 5.1eV, the stopping potential is estimated to be
Solution
To solve this problem, we need to use the formula for the energy difference between two levels in a hydrogen atom, which is given by:
ΔE = 13.6 * (1/n1² - 1/n2²) eV
where n1 and n2 are the principal quantum numbers of the two levels, and 13.6 eV is the ionization energy of hydrogen.
In this case, the electron jumps from n=3 to n=1. So, we substitute these values into the formula to get:
ΔE = 13.6 * (1/1² - 1/3²) = 13.6 * (1 - 1/9) = 13.6 * 8/9 = 12.09 eV
This is the energy of the photons emitted by the hydrogen atom.
Next, we use the photoelectric effect equation, which is given by:
E = φ + KE_max
where E is the energy of the incident photons, φ is the work function of the material, and KE_max is the maximum kinetic energy of the ejected electrons. The stopping potential V is related to the maximum kinetic energy by the equation KE_max = eV, where e is the charge of an electron.
Rearranging the photoelectric effect equation for V gives:
V = (E - φ) / e
Substituting the given values gives:
V = (12.09 eV - 5.1 eV) / e = 6.99 V
So, the stopping potential is estimated to be 6.99 V.
Similar Questions
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