HW Problem A15


Moderators: Chem_Mod, Chem_Admin

Hannah Lee 2F
Posts: 105
Joined: Thu Jul 11, 2019 12:15 am

HW Problem A15

Postby Hannah Lee 2F » Fri Oct 11, 2019 1:56 pm

1A.15) In the ultraviolet spectrum of atomic hydrogen, a line is observed at 102.6 nm. Determine the values of n for the initial and final energy levels of the electron during the emission of energy that leads to this spectral line.

Is there any way to solve this without the Rydberg equation, using the way we solved it in class, by using En = final - En = initial? Or would we need to use the Rydberg equation for this unit?

005391550
Posts: 92
Joined: Thu Jul 25, 2019 12:15 am

Re: HW Problem A15

Postby 005391550 » Fri Oct 11, 2019 2:09 pm

if you derived En(final) - En(initial) it would do the same thing as the Rydberg equation since En = -hR/n^2. so technically you can do either and still get the same result!

Angus Wu_4G
Posts: 95
Joined: Fri Aug 02, 2019 12:15 am

Re: HW Problem A15

Postby Angus Wu_4G » Fri Oct 11, 2019 3:25 pm

I answered this question in a different discussion thread, but I'll repost my explanation below again:

You would first use the wavelength you have and find the energy change using E=(hc)/wavelength.

Once you have the energy change, use the equation E=(-hR)/n^2 with n=1 to find the energy of the electron at n=1.

Once you have the energy change and the energy of the electron at n=1, you can set up this equation:

(Energy of electron at n=final) - (Energy of electron at n=1) = Energy change

Then finally, solve for n final. According to the solution manual, n should be 3.

As for the Rydberg equation, it is the same as the equation that was given to us in lecture, you could use either one, both will yield the same answer. If you do use the Rydberg Equation though, you have to be careful with your n initial and n final values and substitute then correctly or else you might get a negative energy.


Return to “Properties of Light”

Who is online

Users browsing this forum: No registered users and 1 guest