An electron in a hydrogen atom undergoes a transistion from the n=5 state to ano

Question

An electron in a hydrogen atom undergoes a transistion from the n=5 state to another state by emitting a photon.

A) What are the possible states that this electron after the transistion, and calculate the energy of the photon that is emitted for each of the possible transistions.

b) what are the final staes of the electron if the photon is in the visible part of the Em spectrum(roughly 1.9-3.2 eV) Calculate the frequency, wavelenght, and photons emitted.

c)Suppose that the photons that you identified in part a are incident on a metal having a work function phi=1.51eV. Is it possible for an electron to ejected from the surface of this metal? Calculate the maximum kinetic energy of the electrons that are ejected and explain why the other photons do not emit electron.

Explanation / Answer

Recall, the Rydberg formula for the energy of the emitted photon that undergoes a transition between two energy levels.

E(photon) = Eo[1 / (n1)^2 - 1/(n2)^2]
where n1<n2

We also have the Plank/Einstein relation
E = h f
where E = photon energy (J); h = Plank's constant (Js); f = frequency of photon (Hz)

So lets have a go at getting the energy of the emitted photon.

Using the Rydberg formula
E(emitted photon) = - 13.6 ( 1/9 - 1/100) = 1.375 eV

To get the frequency we need to convert the eV into joules. So
1.375 * 1.6 * 10^-19 = 2.2 * 10^-19 J

From the Einstein/Plank relation we get
f = E/h
f = 2.2 * 10^-19 / 6.63 * 10^-34 = 0.331 * 10^15 = 3.31 * 10^14 Hz

From wave mechanics we have
v = f ?
? = v/f = 3.0 * 10^8 / 3.31 * 10^14 = 9.04 * 10^-6 m
? = 9.04 * 10^3 nm

So, to summarise,
a) Wavelength = 9.04 * 10^3 nm
b) Frequency = 3.31 * 10^14 Hz
c) Energy of emitted electron = 1.375 eV

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