Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
55 views
in Chemistry by (80.8k points)
closed by
The frequency of light emitted for the transition `n = 4` to `n =2` of `He^+` is equal to the transition in `H` atom corresponding to which of the following ?
A. `n=2` to `n=1`
B. `n=3` to `n=2`
C. `n=4` to `n=3`
D. `n=3` to `n=1`

1 Answer

0 votes
by (81.1k points)
selected by
 
Best answer
Correct Answer - 1
`hv=DeltaE=13.6 z^(2)(1/n_(1)^(2)-1/n_(2)^(2))rArr v_(He^(+))=v_(H)xxz^(2)(1/((n_(1)/2)^(2))-1/((n_(2)/2)^(2)))=v_(H)(1/((2/2)^(2))-1/((4/2)^(2)))`
For H-atom `rArr n_(1)=1, n_(2)=2`

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...