Question 1
True or False: Bohr postulated that electrons move in welldefined circular
orbits around the nucleus of a hydrogen atom.
A. True 
B. False 
Question 2
In Bohr's model for the H atom, the visible series of spectral lines correspond
to electrons "jumping" down to
A. the smallest orbit from larger orbits. 
B. the second smallest orbit from larger orbits. 
C. the third smallest orbit from larger orbits. 
D. nucleus from one of the orbits. 

Question 3
Using the Bohr model, calculate the energy of the electron in the n=7 orbit.
Question 4
Using the Bohr model, calculate the energy of the photon produced when the
electron in the n=8 orbit "jumps" down to the n=4 orbit.
Question 5
Using the Bohr model, calculate the frequency of light produced when the
electron jumps from the n=6 orbit to the n=2 orbit.
Question 6
Using the Bohr model, calculate the wavelength of light produced when the
electron "jumps" from the n=5 orbit to the n=4 orbit.
Question 1 (00000001A0805100, Variation No. 70): A.
True
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For more information on Bohr's model, click here.
Question 2 (00000001A0805101, Variation No. 53): B.
the second smallest orbit from larger orbits.
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The visible series of lines in the H atom spectrum is part of the Balmer
series. For more information on Bohr's model, click here.
Question 3 (00000001A0805400, Variation No. 57): 4.45x10^{20}
J
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Orbits in the Bohr model are identified by the quantum number n. The
first allowed orbit is n=1, the next allowed orbit is n=2, etc. The energy of
the electron in an orbit was derived by Bohr to be equal to:
E = R/n^{2}
where the theoretical value of the constant R happens to be equal to the
experimentally known value of the Rydberg constant (2.18x10^{18} J).
Thus, the energy of the electron in the n=7 orbit is
E = (2.18x10^{18} J) / 7^{2} = 4.45x10^{20} J
NOTE:
1. The energy is negative because it is in comparison to the case where the
nucleus and electron are infinitely separated (which is defined to have zero
energy). In other words, for an electron and proton, initially separated by
infinite distance, to come together to form a hydrogen atom, energy must be
given up. Conversely, for an electron in the first orbit to be infinitely
separated from the nucleus, it must absorb 2.18x10^{18} J of energy;
we say that this is the ionization energy of Hydrogen.
2. Energies of electrons in atoms are often quoted in eV (electron volt); 1 eV
= 1.602x10^{19} J. The Rydberg constant in eV is 13.6 eV. The
ionization energy of hydrogen is 13.6 eV.
Question 4 (00000001A0805402, Variation No. 28): 1.02x10^{19}
J
Feedback
Orbits in the Bohr model are identified by the quantum number n. The
first allowed orbit is n=1, the next allowed orbit is n=2, etc. The energy of
the electron in an orbit was derived by Bohr to be equal to:
E = R/n^{2}
where the theoretical value of the constant R happens to be equal to the
experimentally known value of the Rydberg constant (2.18x10^{18} J).
Thus, the energy of the electron in the n=4 orbit is
E_{4} = (2.18x10^{18} J) / 4^{2} = 1.362x10^{19}
J
The energy of the electron in the n=8 orbit is
E_{8} = (2.18x10^{18} J) / 8^{2} = 3.406x10^{20}
J
The energy of the photon produced is equal to the energy lost by the electron
as it jumps to the lower orbit.
E(photon) = E_{8}  E_{4}
= (1.362x10^{19} J)  (3.406x10^{20} J) = 1.02x10^{19}
J
Question 5 (00000001A0805405, Variation No. 30): 7.30x10^{14}
Hz
Feedback
Orbits in the Bohr model are identified by the quantum number n. The
first allowed orbit is n=1, the next allowed orbit is n=2, etc. The energy of
the electron in an orbit was derived by Bohr to be equal to:
E = R/n^{2}
where the theoretical value of the constant R happens to be equal to the
experimentally known value of the Rydberg constant (2.18x10^{18} J).
Thus, the energy of the electron in the n=2 orbit is
E_{2} = (2.18x10^{18} J) / 2^{2} = 5.450x10^{19}
J
The energy of the electron in the n=6 orbit is
E_{6} = (2.18x10^{18} J) / 6^{2} = 6.056x10^{20}
J
The energy of the photon produced matches the energy lost by the electron,
which is equal to difference between the two orbits.
E(photon) = E_{6}  E_{2}
= (5.450x10^{19} J)  (6.056x10^{20} J) = 4.84x10^{19}
J
The energy of a photon is related to the frequency ()
by:
E(photon) = h,
where h=Planck's constant=6.626x10^{34} J s
Thus:
=
E(photon)/h = 4.84x10^{19} J / 6.626x10^{34} J s = 7.30x10^{14}
s^{1}, or 7.30x10^{14} Hz
Question 6 (00000001A0805406, Variation No. 77): 4.05x10^{3}
nm
Feedback
Orbits in the Bohr model are identified by the quantum number n. The
first allowed orbit is n=1, the next allowed orbit is n=2, etc. The energy of
the electron in an orbit was derived by Bohr to be equal to:
E = R/n^{2}
where the theoretical value of the constant R happens to be equal to the
experimentally known value of the Rydberg constant (2.18x10^{18} J).
Thus, the energy of the electron in the n=4 orbit is
E_{4} = (2.18x10^{18} J) / 4^{2} = 1.362x10^{19}
J
The energy of the electron in the n=5 orbit is
E_{5} = (2.18x10^{18} J) / 5^{2} = 8.720x10^{20}
J
The energy of the photon produced is equal to the energy lost by the electron,
which is equal to the energy difference between the two orbits.
E(photon) = E_{5}  E_{4}
= (1.362x10^{19} J)  (8.720x10^{20} J) = 4.90x10^{20}
J
The energy of a photon is related to the wavelength ()
by:
E(photon) = hc/,
where
h = Planck's constant = 6.626x10^{34} J s,
c=speed of light = 2.998x10^{8} m/s
Thus:
=
hc/E(photon) = (6.626x10^{34} J s)(2.998x10^{8} m/s)/(4.90x10^{20}
J)
= 4.05x10^{6} m, or 4.05x10^{3} nm