A gas absorbs photon of 355 nm and emits at two wavelengths. If one of the emission is at 680 nm, the other is at

Options

A. 1035m
B. 325m
C. 743m
D. 518m

Answer

To solve the problem, we need to use the relationship between the energy of the absorbed and emitted photons and their wavelengths. Here’s a step-by-step solution:

### Step 1: Understand the relationship between energy and wavelength
The energy (E) of a photon is inversely related to its wavelength (λ) and is given by the equation:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h \) is Planck's constant (6.626 x 10^-34 J·s),
- \( c \) is the speed of light (3.00 x 10^8 m/s),
- \( \lambda \) is the wavelength in meters.

### Step 2: Set up the equation for absorbed and emitted energy
According to the conservation of energy, the energy absorbed by the gas must equal the total energy emitted. Thus, we can write:
\[ E_{absorbed} = E_{emitted1} + E_{emitted2} \]

### Step 3: Substitute the energy equation in terms of wavelength
Using the relationship between energy and wavelength, we can express the equation as:
\[ \frac{hc}{\lambda_t} = \frac{hc}{\lambda_1} + \frac{hc}{\lambda_2} \]
Since \( hc \) is a common factor, we can simplify this to:
\[ \frac{1}{\lambda_t} = \frac{1}{\lambda_1} + \frac{1}{\lambda_2} \]

### Step 4: Plug in the known values
From the problem, we know:
- \( \lambda_t = 355 \) nm (the wavelength of the absorbed photon),
- \( \lambda_1 = 680 \) nm (one of the emitted wavelengths).

Now we can substitute these values into the equation:
\[ \frac{1}{355} = \frac{1}{680} + \frac{1}{\lambda_2} \]

### Step 5: Solve for \( \lambda_2 \)
Rearranging the equation to isolate \( \frac{1}{\lambda_2} \):
\[ \frac{1}{\lambda_2} = \frac{1}{355} - \frac{1}{680} \]

Now, we need to calculate the right side:
1. Find a common denominator for the fractions, which is \( 355 \times 680 \).
2. Calculate:
\[ \frac{1}{355} = \frac{680}{241400} \]
\[ \frac{1}{680} = \frac{355}{241400} \]
Thus,
\[ \frac{1}{\lambda_2} = \frac{680 - 355}{241400} = \frac{325}{241400} \]

3. Therefore,
\[ \lambda_2 = \frac{241400}{325} \approx 743 \text{ nm} \]

### Final Answer
The other emitted wavelength \( \lambda_2 \) is approximately **743 nm**.

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