Question

Find the energy (in eV) of a ground state electron in a He II (He + ) atom. (b.) Find the amount of energy (in eV) needed to ionize the ground state electron of part (a). (c.) For which ionization state(s) of Carbon could you calculate (principle) quantum energy levels for given the formula for E n presented in class

Answer 1

Answer:

a)  -54.42 eV

b)  C VI (C +++++)

Explanation:

Hi!

I imagine that the formula given in class is the energy of the hydorgen like atoms, which is:

$E_{n}=-\frac{\text{ \mu e}^4 Z^2}{32 \pi ^2 n^2 \epsilon _0^2 \hbar }$

Where all are physical constants, such as vacuum permitivity ε0, planck's constant (reduced) h bar, electron's charge e, the atomic number Z, and the reduced mass μ.

The reduced mass is given by the formula:

$\frac{1}{\frac{1}{m_{\text{electron}}}+\frac{1}{m_{\text{nucleous}}}}$

In general the electron's mass is much smaller than the nucleous mass therefore the reduced mass is almost equal to the electron's mass.

If we replace all these constants the energy will become:

$E_{n}=-\frac{Z^{2}13.6 eV}{n^{2} }$

For He --> Z = 2, therefore:

$E_{n}=-\frac{54.42 eV}{n^{2} }$

And the ground state is given  when n=1

a)

E = -54.42 eV

b)

This formula applies only to hydorgen like atoms, that is whit only one electron, therefore if an element were to have Z electrons, the formula will only apply to that element ionized Z-1 times.

For C Z=6

We could only calculate the quantum energy levels with the given formula for C VI (C +++++)