Modern Physics 1 Question 19
16. 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?
(a) 2 photons of energy $10.2 eV$
$(2005,2 M)$
(b) 2 photons of energy $1.4 eV$
(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 energy $1.4 eV$
Show Answer
Solution:
- The first photon will excite the hydrogen atom (in ground state) to first excited state (as $E _2-E _1=10.2 eV$ ). Hence, during de-excitation a photon of $10.2 eV$ will be released. The second photon of energy $15 eV$ can ionise the atom. Hence, the balance energy i.e. $(15-13.6) eV=1.4 eV$ is retained by the electron. Therefore, by the second photon an electron of energy $1.4 eV$ will be released.
