The number of photons of light of `barv =2.5×10^6m^(−1)` necessary to provide 1 J of energy are:

Options

A. 2 x `10^14`
B. 2 x `10^18`
C. 1 x `10^14`
D. 3 x `10^18`

Explanation

To solve the problem of finding the number of photons of light necessary to provide 1 J of energy given a wave number of \( \bar{v} = 2.5 \times 10^6 \, \text{m}^{-1} \), 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 \( h \) is Planck's constant and \( \nu \) is the frequency of the light. The frequency \( \nu \) can be related to the wave number \( \bar{v} \) as:
\[ \nu = c \cdot \bar{v} \]
where \( c \) is the speed of light.

### Step 2: Substitute the wave number into the energy equation
We can rewrite the energy equation in terms of wave number:
\[ E = n \cdot h \cdot (c \cdot \bar{v}) \]
Rearranging gives us:
\[ n = \frac{E}{h \cdot c \cdot \bar{v}} \]

### Step 3: Plug in the known values
We know:
- \( E = 1 \, \text{J} \)
- \( h = 6.626 \times 10^{-34} \, \text{J s} \) (Planck's constant)
- \( c = 3.00 \times 10^8 \, \text{m/s} \) (speed of light)
- \( \bar{v} = 2.5 \times 10^6 \, \text{m}^{-1} \)

Substituting these values into the equation for \( n \):
\[ n = \frac{1}{(6.626 \times 10^{-34}) \cdot (3.00 \times 10^8) \cdot (2.5 \times 10^6)} \]

### Step 4: Calculate \( n \)
Now, we perform the calculation:
1. Calculate \( h \cdot c \cdot \bar{v} \):
\[ h \cdot c \cdot \bar{v} = (6.626 \times 10^{-34}) \cdot (3.00 \times 10^8) \cdot (2.5 \times 10^6) \]
\[ = 4.9695 \times 10^{-19} \, \text{J} \]

2. Now calculate \( n \):
\[ n = \frac{1}{4.9695 \times 10^{-19}} \approx 2.014 \times 10^{18} \]

### Step 5: Final answer
Thus, the number of photons necessary to provide 1 J of energy is approximately:
\[ n \approx 2.014 \times 10^{18} \text{ photons} \]

This can be rounded to:
\[ n \approx 2 \times 10^{18} \text{ photons} \]