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1) The ionization energy of the hydrogen atom is the energy required to remove i

ID: 901865 • Letter: 1

Question

1) The ionization energy of the hydrogen atom is the energy required to remove its electron. Calculate the ionization energy of hydrogen with its electron in the first orbit. Then calculate the ionization energy for the hydrogen atom with its electron in the second orbit.

2) The eletronic transitions with the electron in the fourth orbit make up the brackett series of lines in the emission spectrum. Calculate the wavelength of the lowest energy light and the wavelength of the highest energy light belonging to this series.

3) Hydrogen atoms absorb energy so that electrons are excited up to the n=7 energy level. Electrons then udergo these transitions, among others: (a) n=7 to n=1; (b) n=7 to n=6; and (c) n=2 to n=1. Which transition produces a photon with (i) the smallest energy; (ii) the highest frenquency; (iii) the shortest wavelength?

4)Radio waves are electromagnetic waves. Calculate the wavelength (m) and energy (J) of an FM radio wave with frequency of 100 megaertz.

5) There are many allowed orbits for the electron in Born's hydrogen atom. Are these prbits closer together for large valeus of n or for small values of n?

6)Are energies of orbits closer together for larges values of n or for small value of n?

Explanation / Answer

To Calculate the ionization energy of hydrogen with its electron in the first orbit use the following equation:

E= -2.178 x 10^-18 (Z^2/n^2)

for energy of a valence level n =1 and Z for hydrogen = 1



E= -2.178 x 10^-18 (1^2/1^2)

E= -2.178 x 10^-18 J

E= -2.178 x 10^-18 J per 1 atom

To Calculate the ionization energy of hydrogen with its electron in the second orbit use the following equation:

E= -2.178 x 10^-18 (Z^2/n^2)

for energy of a valence level n =2 and Z for hydrogen = 1



E= -2.178 x 10^-18 (1^2/2^2)

E= -0.5445 x 10^-18 J

E= -0.5445 x 10^-18 J per 1 atom