Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9

Q.

A hydrogen atom moves with a velocity u and makes a head on inelastic collision with another stationary hydrogen atom. Both are in the ground state before collision. What is the minimum value of u (in x104 m/s ), if one of them is to be given a minimum excitation energy? The ionization energy is 13.6 eV. Mass of the hydrogen atom is 1.0078 x 1.66 x 10-27 kg.

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya

answer is 0006.24.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Detailed Solution

From conservation of momentum mu=2mv⇒ v=u/2..........(i) The energy of excitation ΔE is given by ΔE=12mu2−2×12m(u/2)2⇒ ΔE=14mu2.......(ii)The minimum excitation energy for a H-atom is for transition n1 = 1 to n2 = 2 which corresponds to an energy of 10.2 eVFrom Eq. (2)14×1.0078×1.66×10−27u2=10.2×1.6×10−19⇒u=6.24×104m/s
Watch 3-min video & get full concept clarity
score_test_img

courses

No courses found

Get Expert Academic Guidance – Connect with a Counselor Today!

best study material, now at your finger tips!

  • promsvg

    live classes

  • promsvg

    progress tracking

  • promsvg

    24x7 mentored guidance

  • promsvg

    study plan analysis

download the app

gplay
mentor

Download the App

gplay
whats app icon
personalised 1:1 online tutoring