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If an electron makes a transition from the n = 4 to the n = 1
Bohr orbital in a hydrogen atom, determine the wavelength of
the light emitted and the recoil speed of the atom.
Solution:
To determine the wavelength, we use
= R
With R = 1.097 x 10^{7} m, n_{i} = 4 , and n_{f} = 1 , we find = 97.2 nm. This photon carries away momentum given byp = = = 6.62 x 10^{ 27} ,
which by momentum conservation must be the recoil momentum of the atom. Using the mass of the hydrogen atom as 1.67 x 10^{ 27} kg, we find it's speed must bev = = = 4.1 .
Thus, the wavelength of the emitted photon is 97.2 nm, while the recoil speed of the hydrogen atom is 4.1 m/s.
A photon incident on a hydrogen atom causes the electron to make a transition
from the n = 1 orbital to the n = 3 orbital. What is the wavelength of the
photon, and what are the possible wavelengths of the emitted radiation
when the electron returns to the n = 1 state?
Solution:
We assume the photon gives up all of its energy to the electron. In making
a transition from n = 1 to n = 3 , the electron gains an energy
E =  13.6 eV  13.6 eV 12.1 eV ,
which by energy conservation must be equal to the energy of the incident photon. The associated wavelength is then found using= = = 102.8 nm .
When the electron in the excited n = 3 state falls back to the ground state with n = 1 , it can do so by 2 distinct paths: 31 and 321 . This would correspond to 3 distinct wavelengths of the emission spectra.Thus, the wavelength of the incident photon is 102.8 nm, and there would be 3 possible wavelengths present in the emitted radiation when the electron returned to the n = 1 state.
An electron is in the first Bohr orbital of the hydrogen
atom. Determine
the speed of the electron, the time it takes to complete one complete
revolution, and the current in Ampères corresponding to this moving charge.
Solution:
The energy of the electron is given by
E =  =  ,
which in the first Bohr orbital with n = 1 corresponds to an energy of  13.6 eV. Using the relations= mv^{ 2},
we find this corresponds to a kinetic energy ofmv^{ 2} =  E = 13.6 eV x = 2.2 x 10^{ 18} J .
From this we obtain the speed asv = = = 2.2 x 10^{6} .
To obtain the time taken to complete one revolution we divide the distance travelled ( = 2a_{0} = 2 x 0.053 nm) by the speed:T = = = 1.52 x 10^{ 16} s .
This corresponds to a current ofI = = = = 1.1 x 10^{ 3} A .
Thus, the speed of the electron in 2.2 x 10^{6} m/s, it takes 1.52 x 10^{ 16} s to complete one revolution, and the associated current is 1.1 x 10^{ 3} A.
Consider a string fixed at both ends which vibrates in a standing wave
pattern, as in Fig. 28.3.
Suppose such a wave pattern describes a free particle of mass m confined to a box of length L . Find the allowable energy levels of such a particle.
Solution:
In order for a standing wave to exist between the ends, we must have
a half integral or else an integral number of wavelengths present:
n = Ln = 1,2,3,...
Using the de Broglie hypothesis, we can relate this wavelength to the momentum of the particle as= = v = .
From this, we find the kinetic energy of the particle to beE_{n} = mv^{ 2} = .
Thus, the energy of the particle is quantized as E = E_{0}n^{ 2}, where E_{0} = h^{ 2}/8mL^{ 2}.
Write out the possible electronic energy states for an electron
in the n = 1 and 2 energy levels.
Solution:
For a given value of n , the orbital quantum number can assume
values
l = 0,1,2,...,n  1 , the orbital magnetic quantum
number can be
m_{l} =  l,  l + 1,...,0,...,l  1,l , and the spin
quantum number can be
m_{s} = .
For n = 1 we have only l = 0 , for which m_{l} = 0 is the only allowed value. Thus, we find
For n = 2 , l can assume values 0 (with m_{l} = 0 ) or 1 (for which m_{l} =  1,0,1 ). We then have
n  l  m_{l}  m_{s} 
2 
 

2 
+ 

2  1   1 
 
2  1   1 
+ 
2  1 
 

2  1 
+ 

2  1  + 1 
 
2  1  + 1  + 
Note that at this point we cannot say which of the various states for a given value of n have lower energy.
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