if The energy difference between the ground state and excited state of an atom is `4.4xx 10^(-19 ) J .` the wavelength of photon required to produce this transition is

Options

A. `4.5xx10 ^(-7) m`
B. ` 4.5 xx 10 ^(-7) nm`
C. ` 4.5 xx 10 ^(-7) Å`
D. `4.5 xx 10 ^(-7 ) cm`

Answer

To find the wavelength of the photon required to produce the transition from the ground state to the excited state of an atom, we can use the relationship between energy and wavelength given by the equation:

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

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 1: Rearrange the equation to solve for wavelength \(\lambda\)

We can rearrange the equation to find the wavelength:

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

### Step 2: Substitute the values into the equation

Now, we can substitute the values for \(h\), \(c\), and \(E\):

\[ \lambda = \frac{(6.626 \times 10^{-34} \, \text{J s}) \times (3.00 \times 10^{8} \, \text{m/s})}{4.4 \times 10^{-19} \, \text{J}} \]

### Step 3: Calculate the numerator

First, calculate the numerator:

\[ 6.626 \times 10^{-34} \times 3.00 \times 10^{8} = 1.9878 \times 10^{-25} \, \text{J m} \]

### Step 4: Calculate the wavelength

Now, divide the result by the energy:

\[ \lambda = \frac{1.9878 \times 10^{-25} \, \text{J m}}{4.4 \times 10^{-19} \, \text{J}} \approx 4.51 \times 10^{-7} \, \text{m} \]

### Step 5: Convert to nanometers

To convert meters to nanometers, we multiply by \(10^{9}\):

\[ \lambda \approx 4.51 \times 10^{-7} \, \text{m} \times 10^{9} \, \text{nm/m} \approx 451 \, \text{nm} \]

### Final Answer

The wavelength of the photon required to produce this transition is approximately **451 nm**.

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