How Does Electron Transition Calculate Atomic Energy Levels and Ionization?

[StephenDoty]

E of nlow to nhigh= -2.178*10^-18 * Z^2/nhigh^2 - nlow^2

find the energy needed to remove an electron completely from a hydrogen atom and the energy needed to remove one mol of electrons from one mol of hydrogen atoms

Z= 1 nhigh= infinity nlow=1

E = 2.178*10^-18 J
one mole of electrons = 2.178*10^-18 * 6.022*10^23 = 1.312*10^6 J/mol
Are these right??

And use the above formula to find the value of Z for an ion whose 2 to 1 transition is associated with a wavelength of 13.4nm.

lamba of 2 to 1= hc/(2.178*10^-18 * Z^2/nhigh^2 - nlow^2)
z=sqrt(hc(nhigh^2-nlow^2)/(13.4*10^-9)(2.178*10^-18))
Z= 4.5
or 5
Is this right??

Thank you guys so much for your help.
Stephen

 

[Borek]

E of nlow to nhigh= -2.178*10^-18 * Z^2/nhigh^2 - nlow^2

Try to use TEX or at least brackets, it doesn't look OK at the moment.

 

[Redbelly98]

Hello Stephen,

Looks good on the single hydrogen atom and mole-of-hydrogen question.

But it looks like there's an error somewhere in the Z question. The correct answer is very close to an integer.

I think the problem is with the expressions you are writing. There should be a term

( 1/nhigh^2 - 1/nlow^2 )​

but instead you have

nhigh^2 - nlow^2​

?

Regards,

Mark

p.s. Borek is correct, it's better to at least use brackets (parantheses) to express things properly and avoid confusion.

 

[StephenDoty]

Z^2/nhigh^2 - nlow^2
is the same thing as z^2*(1/nhigh^2 - 1/nlow^2)

 

[Redbelly98]

Okay, but in your expression

z = sqrt(...)​

it has mysteriously become, literally,

(nhigh^2-nlow^2)​

and that is wrong.

Try keeping it as

(1/nhigh^2 - 1/nlow^2)​

Also, you might find it easier to figure out what the energy is for 13.4 nm, and then work with the energy equation (1st equation of your 1st post).