Energy needed to remove both electrons from a He atom

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1. Dec 26, 2017

Pushoam

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Both electrons are in 1s orbit.

For taking out the second electron, I will have to put slightly more energy than 24.6eV.

So, the energy required to remove both electrons should be slightly more than 49.2 eV.

So, I guess that the option should be (d)

Is this correct?

2. Dec 26, 2017

Dami121

No. It is true that you must put more energy, but actually not so slightly. If you look at this table you can see that the second ionization energy is much higher than the first one.

I think your problem might be that you're confusing the second ionization energy with the total energy required to take both electrons out. https://en.wikipedia.org/wiki/Ionization_energy

3. Dec 26, 2017

Pushoam

The total energy of an electron in n th state for H- like atom is given by :

$E_n = - 13.6 eV ~\frac { Z^2}{n^2}$ .....(1)

I need to provide energy equal and opposite to the potential energy to remove an electron (taking potential energy of electron to be 0 at infinity).

$U_n = - 2 E_n = - 27.2 eV ~\frac { Z^2}{n^2}$ .....(2)

For Helium atom, for both electrons, n = 1.

Now, the ionization potential for the 1st electron could not be determined using (2) as (2) is applicable for One electron atom or ions and He – atom has 2 electrons.

For removal of 2nd electron, (2) could be applied.

Hence, the energy needed for removing the 2nd electron is, 27.2 * 4 eV = 108.8 eV.

So, the minimum energy required to remove both the electrons is (24.6 +108.8) eV = 133.4 eV.

But, this is not given in the options.

What to do now?

4. Dec 26, 2017

Dami121

This is not right. The total energy of an electron in a H- like atom is only:
$E_n = ~\frac {- 13.6 eV}{n^2}$

Think again about what you wrote in (1) and see if you couldn't use it somehow. You're on the right track and really close to the solution.