Need Help Solving Ne9+ Balmer Series Problem

[mirella]

I could use some assistance! - thanks

Question:
Atoms can be ionized by thermal collisions at very high temperatures. One such ion is Ne9+, a neon atom with only a single electron.

(a) By what factor are the energies of its hydrogen-like levels greater than those of hydrogen?

(b) What is the wavelength of the first line in this ion's Balmer series?

(I'm aware of this equation, I'm just not sure how to apply it to this question.)

 

[Andrew Mason]

I could use some assistance! - thanks

Question:
Atoms can be ionized by thermal collisions at very high temperatures. One such ion is Ne9+, a neon atom with only a single electron.

(a) By what factor are the energies of its hydrogen-like levels greater than those of hydrogen?

In the Bohr model, how does the electron's energy depend on the charge of the nucleus (Ze)? What is the Z for Ne9+?

(b) What is the wavelength of the first line in this ion's Balmer series?

Once you get a), b) follows from the Bohr formula.

AM

 

[SpaceTiger]

You need to show some work.

(a) By what factor are the energies of its hydrogen-like levels greater than those of hydrogen?

It seems like you should have a formula for this. If not, perhaps you're expected to use the Bohr model. What is the atomic number of Neon?

(b) What is the wavelength of the first line in this ion's Balmer series?

Which transition does this correspond to?

 

[mirella]

for a) is just the atomic # squared 10^2= 100

for b) I have the equation:

hc/lambda=(Z^2)*(13.6eV) (1/2sq - 1/3sq)

where Z=10

...but it doesn't turn out right, I'm not sure what I'm doing wrong

 

[nrqed]

for a) is just the atomic # squared 10^2= 100

for b) I have the equation:

hc/lambda=(Z^2)*(13.6eV) (1/2sq - 1/3sq)

where Z=10

...but it doesn't turn out right, I'm not sure what I'm doing wrong

What number do you get? What value are you using for h?

 

[mirella]

h = 6.63*10^-34 (a constant right?)

to solve for lambda i get: 1.05*10^-27

 

[nrqed]

h = 6.63*10^-34 (a constant right?)

to solve for lambda i get: 1.05*10^-27

This is what I thought the problem was... you are mixing units! This h in in Joule-second. But the energy you get in in electronvolts. So you have two choices: either you convert the right hand side bacin Joules (by multiplying your energy in eV by 1.602x10^(-19)) or you use the value of h in eV-second.

Pat

 

[mirella]

ahh.. thank you kind sir :)

-ali