# Which electronic transition of the hydrogen spectrum corresponds to "the third line from the red end"?

Question

In Bohr series of lines of hydrogen spectrum, the third line from the red end corresponds to which one of the following inner-orbit jumps of the electron for Bohr orbits in an atom of hydrogen?

(A) $$3\to2$$

(B) $$5\to2$$

(C) $$4\to1$$

(D) $$2\to5$$

Only one option is correct.

My approach

I just drew the electron transition diagram like this:

Then, I counted the third line from the red end, i.e., from the left side starting from the first line of the particular series. I then counter checked with the given options. Surprisingly two options matched, i.e., option B as well as C. I don't know further how to decide which is the correct answer out of B and C, since only one answer is correct. There is no other details mentioned in the question and thus I am confused.

• 4-1 and 5-2 fit the question as you suspect. Aug 15, 2019 at 7:32
• It is kind of an odd question. However since the question mentions a red line, I'd assume that it is referring to the Balmer series which is in the visible range. Thus the answer would be (B).
– MaxW
Aug 15, 2019 at 8:57

The transition $$n_2 \to n_1$$ can be easily calculated by using the following formula and it is not necessary to draw the transitions, which might be difficult for higher transitions where drawing the transitions would be time-consuming.
$$n_2=n_1 + n$$ where $$n$$ represents the $$n^{th}$$ line in a particular series. Since here we are talking about Balmer series, $$n_1=2$$ and for the third line $$n=3$$. Substituting the values $$n_2=2+3=5$$