The first line of Balmer series has wavelength 6563 ˙A. What will be the wavelength of the first line of Lyman series?
We know that,
1λ=R[1n21−1n22]
The first line of Balmer series corresponds to the transition from n2=3→n1=2
⇒1λB=R(122−132)
⇒1λB=R(14−19)
⇒1λB=5R36 .....(1)
Similarly,
The first line of Lyman series corresponds to the transition from n2=2→n1=1
⇒1λL=R(112−122)
⇒1λL=3R4 .....(2)
Using (1) and (2),
⇒1/λB1/λL=(5R36)(3R4)=53×19
⇒λLλB=527
⇒λL=527λB=527×6563 ˙A
⇒λL=1215.4 ˙A
Hence, option (A) is correct.
Why this question? This question is very useful to understand the emission spectra of hydrogen atom, which very important concept for JEE Mains. |