The recoil speed of a hydrogen atom after it emits a photon is going form n=5 state to n =1 state is ….. m/s.

Explanation

To find the recoil speed of a hydrogen atom after it emits a photon while transitioning from the n=5 state to the n=1 state, we can follow these steps:

### Step 1: Calculate the energy difference (ΔE) between the two states.
The energy of an electron in a hydrogen atom can be calculated using the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
For n=5:
\[ E_5 = -\frac{13.6 \, \text{eV}}{5^2} = -\frac{13.6 \, \text{eV}}{25} = -0.544 \, \text{eV} \]

For n=1:
\[ E_1 = -\frac{13.6 \, \text{eV}}{1^2} = -13.6 \, \text{eV} \]

Now, calculate the energy difference:
\[ \Delta E = E_5 - E_1 = (-0.544 \, \text{eV}) - (-13.6 \, \text{eV}) = 13.6 - 0.544 = 13.056 \, \text{eV} \]

### Step 2: Convert ΔE from eV to Joules.
Using the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ \Delta E = 13.056 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 2.09 \times 10^{-18} \, \text{J} \]

### Step 3: Use conservation of momentum to find the recoil speed (V) of the hydrogen atom.
According to the conservation of momentum:
\[ mv = \frac{\Delta E}{c} \]
Where:
- \( m \) is the mass of the hydrogen atom, approximately \( 1.67 \times 10^{-27} \, \text{kg} \)
- \( c \) is the speed of light, approximately \( 3 \times 10^8 \, \text{m/s} \)

Rearranging the equation to solve for \( v \):
\[ v = \frac{\Delta E}{m c} \]

### Step 4: Substitute the values into the equation.
\[ v = \frac{2.09 \times 10^{-18} \, \text{J}}{(1.67 \times 10^{-27} \, \text{kg})(3 \times 10^8 \, \text{m/s})} \]

### Step 5: Calculate the speed.
Calculating the denominator:
\[ (1.67 \times 10^{-27} \, \text{kg})(3 \times 10^8 \, \text{m/s}) = 5.01 \times 10^{-19} \, \text{kg m/s} \]

Now substituting back:
\[ v = \frac{2.09 \times 10^{-18} \, \text{J}}{5.01 \times 10^{-19} \, \text{kg m/s}} \approx 4.17 \, \text{m/s} \]

### Final Answer:
The recoil speed of the hydrogen atom after it emits a photon is approximately **4.17 m/s**.

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