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

Options

A. `4.8xx 10^(-23)kg-m//s`
B. `4.8xx 10^(-27)kg-m//s`
C. `4.8xx 10^(-30)kg-m//s`
D. `7.8xx 10^(-27)kg-m//s`

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 SI units (Joules). The conversion factor is:
1 eV = 1.6 × 10^-19 Joules.

\[ E = 9 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 1.44 \times 10^{-18} \, \text{J} \]

### Step 2: Use the Energy-Momentum Relation
The momentum \( p \) of a photon can be calculated using the relation between energy and momentum:
\[ p = \frac{E}{c} \]
where:
- \( E \) is the energy of the photon,
- \( c \) 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
Perform 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 is:
\[ p = 4.8 \times 10^{-27} \, \text{kg m/s} \]

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