Do you graph the energy levels of a ##1s^1## atom the same as a ##1s^2## atom?

[Danielk010]

Homework Statement

Draw an energy-level diagram showing the lowest four levels of singly ionized helium. Show all
possible transitions from the levels and label each transition with its wavelength.

Relevant Equations

$E_n = \frac {-me^2} {32\pi^2\varepsilon_0^2n^2}$

From the first equation, there are 5 constants, e, $\pi$, $\varepsilon_0$, $n^2$, and 32. The only difference is m, where helium has around four times the mass of hydrogen. What I don't get is if there is a difference between the energy levels of the hydrogen and the ionized hellium? Also I get that helium is ionized in this case so would the atom have an electric configuration of $1s^1$? Thank you for any help.

 

[kuruman]

What I don't get is if there is a difference between the energy levels of the hydrogen and the ionized hellium?

What is $Z$ for hydrogen and helium? How do the energy levels depend on $Z$?

 

[Danielk010]

What is $Z$ for hydrogen and helium? How do the energy levels depend on $Z$?

Z for hydrogen is 1 and Z for helium is 2. The higher the Z, the higher the energy level. I got the equation: $E_n = E_0 * \frac {Z^2} {n^2}$

 

[hutchphd]

What does higher mean here? These levels are usually measured relative to zero at ionization (hencenegative). Also be aware that the mass is the "reduced mass" which changes a little.

 

[Danielk010]

Z for hydrogen is 1 and Z for helium is 2. The higher the Z, the higher the energy level. I got the equation: $E_n =$E = E_0 * \frac {Z^2} {n^2}##

What does higher mean here? These levels are usually measured relative to zero at ionization (hencenegative). Also be aware that the mass is the "reduced mass" which changes a little.

So the $E_0$ value would be increased by the Z, nuclear charge and decreased by n, number of electrons. If $E_0$ is negative then it would be vice versa, I think? Since the hellium is ionized so n = 1 and Z = 2, that means it is $4 * E_0$, right?

 

[kuruman]

If $E_0$ is negative then it would be vice versa, I think?

I don't know what you are trying to say here. Vice versa is used to indicate that the reverse of what you have said is also true. For example 'people can perform some tasks better than machines can and vice versa' means that machines can perform some tasks better than people can.

So are you saying that if $E_0$ is negative, then it would be . . . positive?

 

[Danielk010]

I don't know what you are trying to say here. Vice versa is used to indicate that the reverse of what you have said is also true. For example 'people can perform some tasks better than machines can and vice versa' means that machines can perform some tasks better than people can.

So are you saying that if $E_0$ is negative, then it would be . . . positive?

Sorry for the confusion.

What I meant is that when $E_0$ is negative. If Z increased then $E_0$ would decrease. If n increased, $E_0$ would increase as well.

When $E_0$ is positive, when Z increased, $E_0$ would also increase. When n increased, $E_0$ would decrease.

 

[kuruman]

When the magnitude of a negative number increases, the number decreases because it moves farther away from zero. It is more descriptive to say that if $Z$ is increased, the ground state $E_0$ will lie deeper.

 

[pines-demon]

What is 1s²?

 

[Danielk010]

What is 1s²?

The electronic configuration

 

[Danielk010]

When the magnitude of a negative number increases, the number decreases because it moves farther away from zero. It is more descriptive to say that if $Z$ is increased, the ground state $E_0$ will lie deeper.

Ok thank you

 

[pines-demon]

The electronic configuration

Ah right. I do not get the question, but as for the title, we tend to represent orbitals for a single particle (electron) not for two.

 

[Danielk010]

Ok thank you. I think I got the idea