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
Text Solution
Verified by ExpertsC
The first photon will excite the hydrogen atom (in ground state) to first excited state (as 
). Hence, during de-excitation a photon of 10.2 eV will be released.
The second photon of energy 15 eV can ionize the atom. Hence the balance energy i. e., 
is retained by the electron. Therefore, by the second photon an electron of energy 1.4 eV will be released.
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems