In a hydrogen atom, if the radius of the orbit of the electron is doubled, then its energy will ...
a)Decrease by a factor of 2
b)Remain the same
c)Increase by a factor of 2
d)Actually, it is not possible for the radius of the orbit to double
e)Increase or decrease by a factor not listed
Is it going to increase by a factor of 2?
U = - (1/4 pi eo)e^2 / r
Negative because 0 at infinity and drops as it gets closer
F = (1/4 pi eo) e^2/r^2 = m a = m v^2 /r
m v^2 = (1/4 pi eo) e^2/r
so
KE = (1/2) m v^2 = (1/2) (-U)
the KE is half the magnitude of the potential energy at every radius.
Total energy = (1/2)(1/4 pi eo)e^2/r - (1/4pi eo) e^2/r
= -(1/2) (1/4 pi eo) e^2/r
Energy at R =-(1/2) (1/4 pi eo) e^2/R
Energy at 2R = -(1/2)(1/4 pi eo) e^2/2R
Because of - sign, the total energy is higher at the bigger radius by a factor of 2
The potential energy goes up by more than the kinetic energy goes down.
Yes, the correct answer is c) Increase by a factor of 2.
To understand why, let's look at the scenario of a hydrogen atom. In the hydrogen atom, the electron orbits the nucleus in discrete energy levels, also known as electron shells or orbitals. The energy of the electron in a hydrogen atom is quantized and depends on the radius of its orbit.
According to Bohr's model of the hydrogen atom, the radius of the electron's orbit is directly proportional to the energy of the electron. Specifically, the energy of the electron is inversely proportional to the square of the radius.
If the radius of the electron's orbit is doubled, the energy would be expected to change. Since the energy is inversely proportional to the square of the radius, doubling the radius would result in the energy being reduced to one-fourth of its original value. Conversely, if the radius is halved, the energy would quadruple.
Therefore, if the radius of the electron's orbit is doubled, its energy will increase by a factor of 2.