What is the energy of photons that corresponds to a wave number of `2.5 xx 10^(-5)cm^(-1)`

Options

A. `2.5 xx 10^(-20)erg`
B. `5.1xx 10^(-23)erg`
C. `4.97 xx 10^(-21)erg`
D. `8.5 xx 10^(-2)erg`

Explanation

To find the energy of photons that corresponds to a wave number of \(2.5 \times 10^{-5} \, \text{cm}^{-1}\), we can follow these steps:

### Step 1: Understand the relationship between energy, wave number, and constants
The energy \(E\) of a photon is given by the formula:
\[ E = h \nu \]
where \(h\) is Planck's constant and \(\nu\) is the frequency of the photon. The frequency can be related to the wave number \(\bar{\nu}\) (in \(\text{cm}^{-1}\)) using the equation:
\[ \nu = c \cdot \bar{\nu} \]
where \(c\) is the speed of light.

### Step 2: Substitute the wave number into the energy formula
We can substitute the expression for frequency into the energy equation:
\[ E = h c \bar{\nu} \]

### Step 3: Convert units
Given that the wave number is in \(\text{cm}^{-1}\), we need to convert it to \(\text{m}^{-1}\) for consistency in SI units:
\[ \bar{\nu} = 2.5 \times 10^{-5} \, \text{cm}^{-1} = 2.5 \times 10^{-5} \times 10^2 \, \text{m}^{-1} = 2.5 \times 10^{-3} \, \text{m}^{-1} \]

### Step 4: Use known constants
Now we can use the values of Planck's constant and the speed of light:
- Planck's constant \(h = 6.626 \times 10^{-34} \, \text{J s}\)
- Speed of light \(c = 3.00 \times 10^8 \, \text{m/s}\)

### Step 5: Calculate the energy
Now substitute the values into the energy formula:
\[ E = (6.626 \times 10^{-34} \, \text{J s}) \times (3.00 \times 10^8 \, \text{m/s}) \times (2.5 \times 10^{-3} \, \text{m}^{-1}) \]

Calculating this step-by-step:
1. Calculate \(h \cdot c\):
\[ h \cdot c = 6.626 \times 10^{-34} \times 3.00 \times 10^8 = 1.9878 \times 10^{-25} \, \text{J m} \]
2. Now multiply by the wave number:
\[ E = 1.9878 \times 10^{-25} \times 2.5 \times 10^{-3} = 4.9695 \times 10^{-28} \, \text{J} \]

### Step 6: Convert to a more convenient form
To express this in scientific notation:
\[ E \approx 4.97 \times 10^{-28} \, \text{J} \]

### Final Answer
The energy of the photons that corresponds to a wave number of \(2.5 \times 10^{-5} \, \text{cm}^{-1}\) is approximately:
\[ E \approx 4.97 \times 10^{-28} \, \text{J} \]