JEE Main 20199 AprilMorning Shift
- Exam
- JEE Main
- Year
- 2019
- Date
- 9 April
- Shift
- Morning Shift
- Subject
- Chemistry
- Chapter
- Structure of Atom
JEE Main 2019 · 9 April · Morning Shift · Chemistry · Structure of Atom
For any given series of spectral lines of atomic hydrogen, let Δ -...
For any given series of spectral lines of atomic hydrogen, let \Delta\overset{-}{v} = {\overset{-}{v}}_\max - {\overset{-}{v}}_\min be the difference in maximum and minimum wave number in .
The ratio is
Options
- A
- B
- C
- D
Correct answer
D.
Step-by-step solution
= \dfrac{{\left({v}_{\text{ max }} - {v}_{\text{ min }}\right)}_{\text{ lyman }}}{{\left({v}_\max - {v}_\min\right)}_{Balmar}}
= \dfrac{{\left({E}_\max - {E}_\min\right)}_{lyman}}{{\left({E}_\max - {E}_\min\right)}_{Balmar}}
,