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: Understand the relationship between energy and momentum of a photon
The energy (E) of a photon is related to its momentum (P) and the speed of light (c) by the equation:
\[ P = \frac{E}{c} \]

### Step 2: Convert energy from electron volts to joules
The energy given is in electron volts (eV). We need to convert this to joules (J) using the conversion factor:
\[ 1 \text{ eV} = 1.6 \times 10^{-19} \text{ J} \]
Thus, for 9 eV:
\[ E = 9 \text{ eV} \times 1.6 \times 10^{-19} \text{ J/eV} = 1.44 \times 10^{-18} \text{ J} \]

### Step 3: Use the speed of light
The speed of light (c) is approximately:
\[ c = 3 \times 10^8 \text{ m/s} \]

### Step 4: Calculate the momentum
Now we can substitute the values of energy and speed of light into the momentum formula:
\[ P = \frac{E}{c} = \frac{1.44 \times 10^{-18} \text{ J}}{3 \times 10^8 \text{ m/s}} \]

### Step 5: Perform the calculation
Calculating the above expression:
\[ P = \frac{1.44 \times 10^{-18}}{3 \times 10^8} \]
\[ P = 4.8 \times 10^{-27} \text{ kg m/s} \]

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

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