A hydrogen atom is in `5^("in")` excited state. When the electron jumps to ground state the velocity of recoiling hydrogen atom is `"…................"` `m//s` and the energy of the photon is `"…........"` eV.

Answer

To solve the problem step by step, we need to determine two things: the energy of the emitted photon when the electron jumps from the 5th excited state to the ground state, and the recoil velocity of the hydrogen atom after the photon emission.

### Step 1: Determine the energy of the emitted photon

The energy levels of a hydrogen atom are given by the formula:

\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]

Where \( n \) is the principal quantum number. For the 5th excited state, \( n = 6 \) (since the ground state is \( n = 1 \)).

1. Calculate the energy of the electron in the 6th state:
\[ E_6 = -\frac{13.6 \, \text{eV}}{6^2} = -\frac{13.6 \, \text{eV}}{36} = -0.3778 \, \text{eV} \]

2. Calculate the energy of the electron in the ground state:
\[ E_1 = -\frac{13.6 \, \text{eV}}{1^2} = -13.6 \, \text{eV} \]

3. The energy of the emitted photon when the electron transitions from \( n = 6 \) to \( n = 1 \) is given by:
\[ E_{\text{photon}} = E_1 - E_6 = (-13.6) - (-0.3778) = -13.6 + 0.3778 = -13.2222 \, \text{eV} \]

Thus, the energy of the photon emitted is approximately \( 13.2 \, \text{eV} \).

### Step 2: Calculate the recoil velocity of the hydrogen atom

1. The energy of the emitted photon can also be expressed in terms of momentum:
\[ E = pc \]
where \( p \) is the momentum of the photon and \( c \) is the speed of light.

2. The momentum of the photon can be calculated using:
\[ p = \frac{E_{\text{photon}}}{c} \]

3. Convert the energy from eV to Joules:
\[ E_{\text{photon}} = 13.2 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 2.112 \times 10^{-18} \, \text{J} \]

4. Now calculate the momentum of the photon:
\[ p = \frac{2.112 \times 10^{-18} \, \text{J}}{3 \times 10^8 \, \text{m/s}} = 7.04 \times 10^{-27} \, \text{kg m/s} \]

5. According to the law of conservation of momentum, the momentum of the recoiling hydrogen atom will be equal in magnitude to that of the emitted photon:
\[ p_{\text{hydrogen}} = m_{\text{H}} v_{\text{recoil}} \]
where \( m_{\text{H}} \) is the mass of the hydrogen atom (\( 1.67 \times 10^{-27} \, \text{kg} \)).

6. Rearranging gives:
\[ v_{\text{recoil}} = \frac{p_{\text{hydrogen}}}{m_{\text{H}}} = \frac{7.04 \times 10^{-27} \, \text{kg m/s}}{1.67 \times 10^{-27} \, \text{kg}} \approx 4.21 \, \text{m/s} \]

### Final Answers:
- The energy of the photon is approximately **13.2 eV**.
- The recoil velocity of the hydrogen atom is approximately **4.21 m/s**.