Question

The Bohr Model of the hydrogen atom proposed that there were very specific energy states that the electron could be in. These states were called stationary orbits or stationary states. Higher energy states were further from the nucleus. These orbits were thought to be essentially spherical shells in which the electrons orbited at a fixed radius or distance from the nucleus. The smallest orbit is represented by n=1, the next smallest n=2, and so on, where n is a positive integer representing the shell or orbit. What is the radius of the n=2 orbit for Hydrogen (Z=1)? Submit Answer Tries 0/10 Each shell has a very specific energy. Note that the energy of zero is used to represent the level at which an electron becomes unbound from the nucleus and can fly free. The energies for the Bohr orbits are all negative, which means they are all shells in which the electron is bound in orbit around the nucleus. What is the energy for the n=5 Bohr orbit for Hydrogen (Z=1) expressed in Joules? (do not enter units) Submit Answer Tries 0/10

What is the energy for the n=5 Bohr orbit for Hydrogen (Z=1) expressed in eV? (do not enter units) Submit Answer Tries 0/10

The Bohr Model does a good job of calculating the energy levels for ions that are hydrogen-like, meaning they may have more protons in the nucleus, but they only have one electron. Examples would be He+1, Li+2, Be+3, .... What is the radius of the n=2 Bohr orbit for B+4 (Boron, Z=5)? Submit Answer Tries 0/10

What is the energy of the n=4 Bohr orbit for Be+3 (Beryllium, Z=4)? You can use your choice of energy units. Make sure to enter units this time.

Answer 1

a)The radius for the n=2 orbit of the hydrogen comes described by the equation:

$r=\frac{n^{2}.0,0529nm}{Z}=4.0,0529nm=2,21Angstrom$

b)In Joules, the energy of this orbit comes described by:

$E=-\frac{Z^{2}.2,17.10^{-18}J}{n^{2}}=-\frac{2,17.10^{-18}J}{5^{2}}=-8,68.10^{-20}J$

c)Changing the Rydberg constant.

$E=-\frac{Z^{2}.13,6eV}{n^{2}}=-\frac{13,6eV}{5^{2}}=-5,44$

d)Using the formula applied for a):

$r=\frac{n^{2}.0,0529nm}{Z}=\frac{4.0,0529nm}{5}=0,423Angstrom$

e)Applying the formula used for c)

$E=-\frac{Z^{2}.13,6eV}{n^{2}}=-\frac{4^{2}.13,6eV}{4^{2}}=-13,6eV$