Explaining Hydrogen-Like Ions and Quantum Numbers for Li2+

[8614smith]

Homework Statement

(a) Explain the term Hydrogen like ion and determine the quantum numbers l,m for the Li²⁺ ion in the states for which n=1 and n=2.

(b) Explain why for a hydrogen-like ion it is reasonable to use the relations,
$$\mu\approx{m_e}$$, $${E_n}=-13.6\left(\frac{Z^2}{n^2}\right)$$

(c) What are the energies of the n=1 and n=2 states in Li²⁺ ?

Homework Equations

reduced mass

The Attempt at a Solution

(a) A hydrogen like ion is an atom with only 1 electron

(b)$$\frac{{M_N}{M_e}}{{M_N}+{M_e}}$$ I've then said that the $$M_e$$ term on the denominator is negligible and the $$M_N$$ terms will cancel.
This are from my notes but reading over them i don't understand it.

(c)I don't know how to do this, when i put n=1 and n=2 into the equation, using Z=1, i don't get the right answer. The correct answers are -122.4eV for n=1 and -30.6eV for n=2.

any hints??

 

[Matterwave]

b) You have the answer already. The mass of the electron is so small compared to the mass of the nucleus that effectively you just have a particle orbiting an "infinite" mass (i.e. the nucleus won't be affected by the electron much).

c) Z=1 is for Hydrogen, not Li2+. Do you know what the Z stands for in that equation?

 

[8614smith]

b) You have the answer already. The mass of the electron is so small compared to the mass of the nucleus that effectively you just have a particle orbiting an "infinite" mass (i.e. the nucleus won't be affected by the electron much).

c) Z=1 is for Hydrogen, not Li2+. Do you know what the Z stands for in that equation?

Ah yes stupid mistake, (c) is fine then, but (b) i still don't get, what does the reduced mass actually represent?

 

[Matterwave]

When you work with a reduced mass, you effectively reduce a system from 2 particles orbiting the center of mass to a system of 1 particle with the reduced mass orbiting the center of mass. This makes many calculations easier, since you don't have to deal with 2 particles all the time. For the system with one mass significantly more massive than the other mass, the center of mass is very close to the center of the massive object. So, in effect your system is already reduced to 1 particle with the "reduced mass=actual mass of the orbiting particle" orbiting around the center of mass.

 

[8614smith]

When you work with a reduced mass, you effectively reduce a system from 2 particles orbiting the center of mass to a system of 1 particle with the reduced mass orbiting the center of mass. This makes many calculations easier, since you don't have to deal with 2 particles all the time. For the system with one mass significantly more massive than the other mass, the center of mass is very close to the center of the massive object. So, in effect your system is already reduced to 1 particle with the "reduced mass=actual mass of the orbiting particle" orbiting around the center of mass.

Ok thanks, that's cleared it up for me!