The number of photons of light of `barv =1.5×10^7m^(−1)` necessary to provide 2 J of energy are:

Options

A. `6.7xx10^(-17)`
B. `1.2xx10^(-15)`
C. `4.2xx10^(-16)`
D. `1.2xx10^(-15)`

Explanation

To solve the problem of finding the number of photons necessary to provide 2 J of energy with a given wave number, 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.

### Step 2: Relate wave number to wavelength and frequency
The wave number \( \bar{\nu} \) is related to the wavelength \( \lambda \) by:
\[ \bar{\nu} = \frac{1}{\lambda} \]
And the frequency \( \nu \) can be related to the wave number as:
\[ \nu = c \cdot \bar{\nu} \]
where \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)).

### Step 3: Substitute the wave number into the energy equation
We can express the energy in terms of the wave number:
\[ E = n \cdot h \cdot c \cdot \bar{\nu} \]
Rearranging this gives:
\[ n = \frac{E}{h \cdot c \cdot \bar{\nu}} \]

### Step 4: Plug in the values
Given:
- \( E = 2 \, \text{J} \)
- \( h = 6.626 \times 10^{-34} \, \text{J s} \)
- \( c = 3 \times 10^8 \, \text{m/s} \)
- \( \bar{\nu} = 1.5 \times 10^7 \, \text{m}^{-1} \)

Now substituting these values into the equation:
\[ n = \frac{2}{(6.626 \times 10^{-34}) \cdot (3 \times 10^8) \cdot (1.5 \times 10^7)} \]

### Step 5: Calculate the denominator
Calculating the denominator:
\[ (6.626 \times 10^{-34}) \cdot (3 \times 10^8) \cdot (1.5 \times 10^7) = 2.9733 \times 10^{-18} \]

### Step 6: Calculate the number of photons
Now substituting back to find \( n \):
\[ n = \frac{2}{2.9733 \times 10^{-18}} \approx 6.71 \times 10^{17} \]

### Final Answer
Thus, the number of photons necessary to provide 2 J of energy is approximately:
\[ n \approx 6.71 \times 10^{17} \]