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 15eV. What wil be observed by the detector?
(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

Explanation

To solve the problem, we need to analyze the interactions of the photons with the hydrogen atom step by step.

### Step 1: Understand the first photon interaction
- A photon with energy of **10.2 eV** collides with a stationary hydrogen atom in its ground state.
- The ground state of hydrogen has an energy level of **E1 = -13.6 eV**.
- The first photon has enough energy to excite the electron to the first excited state (E2), which is at **-3.4 eV**.
- The energy difference between the ground state and the first excited state is:
\[ E2 - E1 = -3.4 \, \text{eV} - (-13.6 \, \text{eV}) = 10.2 \, \text{eV} \]
- Thus, after this interaction, the hydrogen atom is excited to the first excited state.

### Step 2: Understand the second photon interaction
- After a time interval, another photon with energy **15 eV** collides with the same hydrogen atom (now in the first excited state).
- The ionization energy of hydrogen is **13.6 eV**. This means that to ionize the hydrogen atom, we need at least 13.6 eV.
- The energy available for the electron after ionization will be:
\[ \text{Energy of the photon} - \text{Ionization energy} = 15 \, \text{eV} - 13.6 \, \text{eV} = 1.4 \, \text{eV} \]
- Therefore, the second photon ionizes the hydrogen atom, and the excess energy (1.4 eV) is given to the ejected electron.

### Step 3: Determine the final observation
- After both interactions, the detector will observe:
- One photon of energy **10.2 eV** (from the first interaction).
- One electron with energy **1.4 eV** (from the second interaction).

Thus, the correct answer is **(c) One photon of energy 10.2 eV and an electron of energy 1.4 eV**.