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
+1 vote
11.8k views
in Atomic structure by (84.0k points)
recategorized by

Estimate the difference in energy between 1st and 2nd Bohr's orbit for a hydrogen atom. At what minimum atomic number, a transition from n  = 2 to n = 1 energy level would result in the emission of X-rays with  λ = 3.0 x 10-8 m ? Which hydrogen atom-like species does this atomic number correspond to ?

1 Answer

+1 vote
by (67.4k points)
selected by
 
Best answer

 

For the species emitting X-rays with λ = 3.0 x 10-8 m or 3.0 x 10-6 cm

The atomic number'of such species is 2. Hence, it is He+ because Bohr's theory is applicable for mono-electronic species.

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

...