A photon collides with a stationary hydrogen atom in ground state inelastically. Enorgy of the colliding photon is 10.2 eV. Almost instantaneously. Another photon collidos with same hydrogen atom inelastically with an energy of 15 eV. What will be observed by the detector ?

Options

A. two photons of energy 10.2 eV
B. two photons of energy 1.4 eV
C. One photon of energy 10.2 eV and an electron of energy 1.4 eV
D. one electron having kinetic energy nearly 11.6 eV

Answer

To solve the problem step by step, we will analyze the interactions of the photons with the hydrogen atom in its ground state and then in its excited state.

### Step 1: Determine the energy of the hydrogen atom in the ground state.
The energy of a hydrogen atom in the ground state (n=1) is given by:
\[ E_1 = -13.6 \text{ eV} \]

### Step 2: Calculate the energy after the first photon collision.
The first photon has an energy of 10.2 eV. When this photon collides with the hydrogen atom, it can be absorbed by the atom, causing a transition to a higher energy state. The change in energy (ΔE) is:
\[ \Delta E = 10.2 \text{ eV} \]

The new energy of the hydrogen atom after absorbing the photon is:
\[ E' = E_1 + \Delta E = -13.6 \text{ eV} + 10.2 \text{ eV} = -3.4 \text{ eV} \]

### Step 3: Determine the new energy state of the hydrogen atom.
To find the new energy level (n2) corresponding to -3.4 eV, we use the formula for the energy levels of hydrogen:
\[ E_n = -\frac{13.6}{n^2} \text{ eV} \]
Setting \(E_n = -3.4 \text{ eV}\):
\[ -3.4 = -\frac{13.6}{n^2} \]
Solving for \(n^2\):
\[ n^2 = \frac{13.6}{3.4} = 4 \implies n = 2 \]
Thus, the hydrogen atom transitions from n=1 to n=2.

### Step 4: Calculate the energy of the hydrogen atom in the excited state (n=2).
The energy of the hydrogen atom in the n=2 state is:
\[ E_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} = -3.4 \text{ eV} \]

### Step 5: Calculate the energy after the second photon collision.
The second photon has an energy of 15 eV. The hydrogen atom in the n=2 state can only absorb energy up to the ionization limit, which is 3.4 eV in this case. Therefore, the energy absorbed from the second photon is:
\[ \text{Energy absorbed} = 3.4 \text{ eV} \]

The remaining energy of the second photon after absorption is:
\[ \text{Remaining energy} = 15 \text{ eV} - 3.4 \text{ eV} = 11.6 \text{ eV} \]

### Step 6: Determine what will be observed by the detector.
The detector will observe the remaining energy of the second photon, which is:
\[ \text{Observed energy} = 11.6 \text{ eV} \]

### Final Answer:
The detector will observe a photon with an energy of 11.6 eV.

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