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 the energy from eV to Joules
The energy of the photon is given as 9 eV. We need to convert this energy into Joules using the conversion factor:
1 eV = \(1.6 \times 10^{-19}\) Joules.

\[ E = 9 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 14.4 \times 10^{-19} \, \text{J} \]

### Step 2: Use the formula for momentum of a photon
The momentum \(p\) of a photon can be calculated using the formula:

\[ p = \frac{E}{c} \]

where \(E\) is the energy of the photon and \(c\) is the speed of light in a vacuum, approximately \(3 \times 10^8 \, \text{m/s}\).

### Step 3: Substitute the values into the momentum formula
Now, substituting the values we have:

\[ p = \frac{14.4 \times 10^{-19} \, \text{J}}{3 \times 10^8 \, \text{m/s}} \]

### Step 4: Calculate the momentum
Now, we perform the division:

\[ p = \frac{14.4}{3} \times 10^{-19} \times 10^{-8} = 4.8 \times 10^{-27} \, \text{kg m/s} \]

### 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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