The number of photons of light of `barv =2.5×10^7m^(−1)` necessary to provide 3 J of energy are:

Options

A. `6xx10^(17)`
B. `7xx10^(-17)`
C. `2.4xx10^(18)`
D. `7.7xx10^(-18)`

Explanation

To solve the problem of finding the number of photons of light necessary to provide 3 J of energy, given the wave number \( \bar{\nu} = 2.5 \times 10^7 \, \text{m}^{-1} \), we can follow these steps:

### Step 1: Understand the relationship between energy, number of photons, and wave number
The energy \( E \) of \( n \) photons can be expressed as:
\[ E = n \cdot h \cdot \nu \]
Where:
- \( E \) is the total energy (in joules),
- \( n \) is the number of photons,
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{J s} \)),
- \( \nu \) is the frequency of the light.

Since the wave number \( \bar{\nu} \) is given, we can relate it to the frequency \( \nu \) using the equation:
\[ \nu = c \cdot \bar{\nu} \]
Where \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)).

### Step 2: Calculate the frequency \( \nu \)
Using the wave number:
\[ \nu = c \cdot \bar{\nu} = (3 \times 10^8 \, \text{m/s}) \cdot (2.5 \times 10^7 \, \text{m}^{-1}) \]
Calculating this gives:
\[ \nu = 7.5 \times 10^{15} \, \text{Hz} \]

### Step 3: Substitute values into the energy equation
Now we can substitute \( \nu \) into the energy equation:
\[ E = n \cdot h \cdot \nu \]
Rearranging for \( n \):
\[ n = \frac{E}{h \cdot \nu} \]

### Step 4: Substitute known values
Substituting \( E = 3 \, \text{J} \), \( h = 6.626 \times 10^{-34} \, \text{J s} \), and \( \nu = 7.5 \times 10^{15} \, \text{Hz} \):
\[ n = \frac{3}{(6.626 \times 10^{-34}) \cdot (7.5 \times 10^{15})} \]

### Step 5: Calculate \( n \)
Calculating the denominator:
\[ h \cdot \nu = (6.626 \times 10^{-34}) \cdot (7.5 \times 10^{15}) \approx 4.9695 \times 10^{-18} \, \text{J} \]
Now substituting back:
\[ n = \frac{3}{4.9695 \times 10^{-18}} \approx 6.04 \times 10^{17} \]

### Final Answer
Thus, the number of photons necessary to provide 3 J of energy is approximately:
\[ n \approx 6.04 \times 10^{17} \]