A photon of `300` nm is absorbed by a gas and then emits two photons. One photon has a wavelength `496` nm then the wavelength of second photon in nm is :

Options

A. 959
B. 859
C. 759
D. 659

Explanation

To solve the problem, we need to use the conservation of energy principle for the absorption and emission of photons. Here’s a step-by-step solution:

### Step 1: Calculate the energy of the absorbed photon
The energy of a photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \( E \) is the energy of the photon,
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{J s} \)),
- \( c \) is the speed of light (\( 3.00 \times 10^8 \, \text{m/s} \)),
- \( \lambda \) is the wavelength of the photon in meters.

For the absorbed photon with a wavelength of \( 300 \, \text{nm} \) (which is \( 300 \times 10^{-9} \, \text{m} \)):
\[ E_{absorbed} = \frac{(6.626 \times 10^{-34} \, \text{J s})(3.00 \times 10^8 \, \text{m/s})}{300 \times 10^{-9} \, \text{m}} \]

### Step 2: Calculate the energy of the emitted photon
The energy of the emitted photon with a wavelength of \( 496 \, \text{nm} \) (which is \( 496 \times 10^{-9} \, \text{m} \)) can be calculated similarly:
\[ E_{emitted1} = \frac{hc}{\lambda_1} = \frac{(6.626 \times 10^{-34} \, \text{J s})(3.00 \times 10^8 \, \text{m/s})}{496 \times 10^{-9} \, \text{m}} \]

### Step 3: Use conservation of energy
According to the conservation of energy, the energy of the absorbed photon is equal to the sum of the energies of the emitted photons:
\[ E_{absorbed} = E_{emitted1} + E_{emitted2} \]
Thus, we can express it as:
\[ E_{absorbed} = E_{emitted1} + E_{emitted2} \]

### Step 4: Solve for the second emitted photon
Rearranging the equation gives:
\[ E_{emitted2} = E_{absorbed} - E_{emitted1} \]

### Step 5: Calculate the wavelength of the second emitted photon
Using the energy of the second emitted photon, we can find its wavelength using:
\[ E_{emitted2} = \frac{hc}{\lambda_2} \]
Rearranging gives:
\[ \lambda_2 = \frac{hc}{E_{emitted2}} \]

### Step 6: Substitute and calculate
Now we can substitute the values to find \( \lambda_2 \).

### Final Calculation
1. Calculate \( E_{absorbed} \) using \( \lambda = 300 \, \text{nm} \).
2. Calculate \( E_{emitted1} \) using \( \lambda_1 = 496 \, \text{nm} \).
3. Find \( E_{emitted2} \) using the conservation of energy.
4. Finally, calculate \( \lambda_2 \).