Modern Physics Ques 26
- 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?
$(2005,2 M)$
(a) 2 photons of energy $10.2 $ $eV$
(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 $ $eV$
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Answer:
Correct Answer: 26.(c)
Solution:
Formula:
- 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.
BETA
