The energy difference between ground state of an atom and its excited state is `3xx10^(-19)J`. What is the wavelength of photom required for this radiation ?

Options

A. `6.6xx10^(-34)`m
B. `3xx10^(-8)`m
C. `1.8xx10^(-7)`m
D. `6.6xx10^(-7)`m

Explanation

To find the wavelength of the photon required for the radiation corresponding to the energy difference between the ground state and excited state of an atom, we can use the formula that relates energy (E) to wavelength (λ):

\[ E = \frac{hc}{\lambda} \]

Where:
- \( E \) is the energy difference (in joules),
- \( h \) 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 (in meters).

### Step-by-Step Solution:

1. **Rearrange the formula to solve for wavelength (λ)**:
\[ \lambda = \frac{hc}{E} \]

2. **Substitute the known values into the equation**:
- \( h = 6.626 \times 10^{-34} \, \text{J s} \)
- \( c = 3.00 \times 10^{8} \, \text{m/s} \)
- \( E = 3.00 \times 10^{-19} \, \text{J} \)

Plugging these values into the equation gives:
\[ \lambda = \frac{(6.626 \times 10^{-34} \, \text{J s}) \times (3.00 \times 10^{8} \, \text{m/s})}{3.00 \times 10^{-19} \, \text{J}} \]

3. **Calculate the numerator**:
\[ 6.626 \times 10^{-34} \times 3.00 \times 10^{8} = 1.9878 \times 10^{-25} \, \text{J m} \]

4. **Divide by the energy (E)**:
\[ \lambda = \frac{1.9878 \times 10^{-25} \, \text{J m}}{3.00 \times 10^{-19} \, \text{J}} = 6.626 \times 10^{-7} \, \text{m} \]

5. **Convert to nanometers (1 m = \( 10^{9} \) nm)**:
\[ \lambda = 6.626 \times 10^{-7} \, \text{m} \times 10^{9} \, \text{nm/m} = 662.6 \, \text{nm} \]

### Final Answer:
The wavelength of the photon required for this radiation is approximately **662.6 nm**.

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