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

Q.

A photon collides with a stationary hydrogen atom in ground state inelastically. Energy of the colliding photon is 10.2 eV. After a time interval of the order of micro second another photon collides with same hydrogen atom inelastically with an energy of 15 eV. What will be observed by the detector        

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

a

2 photon of energy 10.2 eV

b

2 photon of energy of 1.4 eV

c

One photon of energy 10.2 eV and an electron of  energy 1.4 eV

d

One photon of energy 10.2 eV and another photon of 1.4 eV

answer is C.

(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

Due to 10.2 eV photon one photon of energy 10.2 eV will be detected.
Due to 15 eV photon the electron will come out of the atom with energy (15 - 13.6) = 1.4 eV

 

Watch 3-min video & get full concept clarity
score_test_img

courses

No courses found

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

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