An isolated hydrogen atom emits a photon of energy 9 eV. Find momentum of the photons

Answer

To find the momentum of a photon emitted by an isolated hydrogen atom with an energy of 9 eV, we can follow these steps:

### Step 1: Convert Energy from eV to Joules
The energy of the photon is given as 9 eV. We need to convert this energy into joules since the standard unit of energy in the SI system is joules.

1 eV = \(1.6 \times 10^{-19}\) joules.

So, the energy in joules (E) is:
\[ E = 9 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 1.44 \times 10^{-18} \, \text{J} \]

### Step 2: Use the Energy-Momentum Relation
The momentum (P) of a photon can be calculated using the relationship between energy and momentum:
\[ P = \frac{E}{c} \]
where:
- \(E\) is the energy of the photon in joules,
- \(c\) is the speed of light in vacuum, approximately \(3 \times 10^8 \, \text{m/s}\).

### Step 3: Substitute the Values
Now we can substitute the values into the equation:
\[ P = \frac{1.44 \times 10^{-18} \, \text{J}}{3 \times 10^8 \, \text{m/s}} \]

### Step 4: Calculate the Momentum
Now, performing the calculation:
\[ P = \frac{1.44 \times 10^{-18}}{3 \times 10^8} = 4.8 \times 10^{-27} \, \text{kg m/s} \]

### Final Answer
The momentum of the photon emitted by the hydrogen atom is:
\[ P = 4.8 \times 10^{-27} \, \text{kg m/s} \]

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