A photon of wavelength 300nm interacts with a stationary hydrogen atom in ground state. During the interaction, whole energy of the photon is transferred to the electron of the atom. State which possibility is correct, (consider, Plank's constant `=4xx10^(-15)` eVs, velocity of light `=3xx10^(8)ms^(-1)` ionisation energy of hydrogen =13.6 eV)

Options

A. Eelctron will be knocked out of the atom
B. Eelctron will go to any excited state of the atom
C. Eelctron will go only to first excited state of the atom
D. Eelctron will keep orbiting in the ground state of the atom

Explanation

To solve the problem, we need to determine the energy of the photon and compare it with the ionization energy of the hydrogen atom and the energy required to excite the electron to the first excited state.

### Step-by-Step Solution:

1. **Calculate the Energy of the Photon:**
The energy \( E \) of a photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h = 4 \times 10^{-15} \) eV·s (Planck's constant)
- \( c = 3 \times 10^{8} \) m/s (speed of light)
- \( \lambda = 300 \) nm = \( 300 \times 10^{-9} \) m (wavelength)

Plugging in the values:
\[ E = \frac{(4 \times 10^{-15} \, \text{eV·s}) \times (3 \times 10^{8} \, \text{m/s})}{300 \times 10^{-9} \, \text{m}} \]

\[ E = \frac{12 \times 10^{-7} \, \text{eV·m}}{300 \times 10^{-9} \, \text{m}} = 4 \, \text{eV} \]

2. **Compare with Ionization Energy:**
The ionization energy of hydrogen is given as 13.6 eV. Since the energy of the photon (4 eV) is less than the ionization energy (13.6 eV), the photon does not have enough energy to ionize the hydrogen atom.

3. **Compare with Excitation Energy:**
The energy required to excite the electron from the ground state (n=1) to the first excited state (n=2) can be calculated as follows:
- Energy of the ground state (n=1): \( E_1 = -13.6 \, \text{eV} \)
- Energy of the first excited state (n=2): \( E_2 = -3.4 \, \text{eV} \)

The energy required to move from n=1 to n=2 is:
\[ E_{excitation} = E_2 - E_1 = (-3.4) - (-13.6) = 10.2 \, \text{eV} \]

Since the energy of the photon (4 eV) is also less than the excitation energy (10.2 eV), the photon cannot excite the electron to the first excited state either.

4. **Conclusion:**
Since the photon does not have enough energy to either ionize the hydrogen atom or excite the electron to the first excited state, the correct possibility is that the electron will remain in the ground state.

### Final Answer:
The correct possibility is that the electron will remain in the ground state of the hydrogen atom.