Hello everyone! So I'm prepping for the Physics GRE and I am using the "Conquering the Physics GRE" Textbook by Yoni Kahn and Adam Anderson. In their chapter on Quantum Mechanics/Atomic Physics, there is this one practice problem that isn't entirely clear to me. It goes as follows:
"The radiation emitted by the 1st excited state of hydrogen as it drops to the ground state is called the Lyman alpha line, and has wavelength 122nm. An excited hydrogen atom is observed radiating light at wavelength 488nm. This most likely results from the transition ___ to ___?
A) n = 2 to n = 1
B) n = 3 to n = 1
C) n = 3 to n = 2
D) n = 4 to n = 1
E) n = 4 to n = 2
So I understand the Rydberg formula as 1/lambda = R*(1/(n_f)^21/(n_i)^2) and the lyman alpha transition is from n=2 to n=1
The book's solution starts with: 488nm/122nm = 4 = (1/(1^2)1/(2^2)) / (1/((n_f)^2) 1/((n_i)^2)) which is an inverted form of Rydberg's frequency form where f is proportional to (1/(n_f)^2) (1/(n_i)^2)
I reduce this to 4 = (3/4)*((n_f)^2  (n_i)^2) but none of the solutions work out for me when I plug in the values. Is there something that I am missing? The book says to "let n_i and n_f to be the initial and final states of the 488nm transition"
then the solution seemingly ignores the math above and says to just double the n values for lyman radiation giving n_i=4 and n_f =2. I'd like to understand the reasoning behind doubling the n values of the lyman radiation as it's best to use accuarate shortcuts on the PGRE but without the math working out, I have a hard time accepting this answer. Can anyone provide some clarity? I'd really appreciate it
I'm sorry for the sloppy entry of the equations. I'm not too sure how to write equations on this forum.
Ryberg's equation

 Posts: 9
 Joined: Sat Dec 27, 2014 10:05 pm
Re: Ryberg's equation
You incorrectly simplified the books formula (the part where you say "i reduced this to..."). The two fractions in the denominator cannot be inverted individually, you must first combine them into one fraction before inverting.daxfeliz wrote:Hello everyone! So I'm prepping for the Physics GRE and I am using the "Conquering the Physics GRE" Textbook by Yoni Kahn and Adam Anderson. In their chapter on Quantum Mechanics/Atomic Physics, there is this one practice problem that isn't entirely clear to me. It goes as follows:
"The radiation emitted by the 1st excited state of hydrogen as it drops to the ground state is called the Lyman alpha line, and has wavelength 122nm. An excited hydrogen atom is observed radiating light at wavelength 488nm. This most likely results from the transition ___ to ___?
A) n = 2 to n = 1
B) n = 3 to n = 1
C) n = 3 to n = 2
D) n = 4 to n = 1
E) n = 4 to n = 2
So I understand the Rydberg formula as 1/lambda = R*(1/(n_f)^21/(n_i)^2) and the lyman alpha transition is from n=2 to n=1
The book's solution starts with: 488nm/122nm = 4 = (1/(1^2)1/(2^2)) / (1/((n_f)^2) 1/((n_i)^2)) which is an inverted form of Rydberg's frequency form where f is proportional to (1/(n_f)^2) (1/(n_i)^2)
I reduce this to 4 = (3/4)*((n_f)^2  (n_i)^2) but none of the solutions work out for me when I plug in the values. Is there something that I am missing? The book says to "let n_i and n_f to be the initial and final states of the 488nm transition"
then the solution seemingly ignores the math above and says to just double the n values for lyman radiation giving n_i=4 and n_f =2. I'd like to understand the reasoning behind doubling the n values of the lyman radiation as it's best to use accuarate shortcuts on the PGRE but without the math working out, I have a hard time accepting this answer. Can anyone provide some clarity? I'd really appreciate it
I'm sorry for the sloppy entry of the equations. I'm not too sure how to write equations on this forum.
On a side note, you need not really proceed any further than the step previous to your simplification error. At this point, trial and error can be used. You know it cannot be option A, because this is the 122nm. Options B and C are also out, since the 3 as one of the states will not give such a nice multiple of the 122nm (i.e. the 488nm). Thus, simple try both D and E to see which one satisfies your equation.