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
77 views
in Chemistry by (80.8k points)
closed by
The dissociation energy of `H_(2)` is `430.53 kJ mol^(-1), `If `H_(2)` is of dissociated by illumination with radiation of wavelength `253.7 nm` , the fraction of the radiant energy which will be converted into ikinetic energy is given by

1 Answer

0 votes
by (81.1k points)
selected by
 
Best answer
Correct Answer - `9(8.68%)`
`E_(H-H)` bond dissocation `=(430.53xx10^(3))/(6.023xx10^(23)) J` per molecule `=7.15xx10^(-19) J` per molecule
`E_("Photon")=(hc)/lambda=(6.626xx10^(-34)xx3.0xx10^(8))/(253.7xx10^(-9))=7.83xx10^(-19) J`
Energy converted into kinetic energy = Energy left after dissociation of bond.
`:.` Energy converted into `KE=(7.83-7.15)xx10^(19) J=0.68xx10^(-19) J`
`:. %` of energy converted into `KE=(0.68xx10^(-19))/(7.83xx10^(-19))xx100=8.68%`

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

...