If the energy difference between the ground state of an atom and in excited state is `4.4xx10^(-14)J`, the wavelength of photon required to produce the transition is:

Options

A. `2.26xx10^(-12)m`
B. `1.13xx10^(-12)m`
C. `4.52xx10^(-16)m`
D. `4.52xx10^(-12)m`

Explanation

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 formula that relates energy and wavelength:

### Step 1: Write down the formula
The energy difference (ΔE) between the two states is given by the equation:
\[ \Delta E = \frac{hc}{\lambda} \]
where:
- \( \Delta E \) = energy difference (in Joules)
- \( h \) = Planck's constant (\(6.63 \times 10^{-34} \, \text{J s}\))
- \( c \) = speed of light (\(3 \times 10^8 \, \text{m/s}\))
- \( \lambda \) = wavelength (in meters)

### Step 2: Rearrange the formula to solve for wavelength
To find the wavelength, we can rearrange the formula:
\[ \lambda = \frac{hc}{\Delta E} \]

### Step 3: Substitute the known values
Substituting the known values into the equation:
\[ \Delta E = 4.4 \times 10^{-14} \, \text{J} \]
\[ \lambda = \frac{(6.63 \times 10^{-34} \, \text{J s})(3 \times 10^8 \, \text{m/s})}{4.4 \times 10^{-14} \, \text{J}} \]

### Step 4: Calculate the numerator
Calculating the numerator:
\[ hc = 6.63 \times 10^{-34} \times 3 \times 10^8 = 1.989 \times 10^{-25} \, \text{J m} \]

### Step 5: Calculate the wavelength
Now substituting back into the equation for wavelength:
\[ \lambda = \frac{1.989 \times 10^{-25}}{4.4 \times 10^{-14}} \]
Calculating this gives:
\[ \lambda = 4.52 \times 10^{-12} \, \text{m} \]

### Step 6: Final answer
Thus, the wavelength of the photon required to produce the transition is:
\[ \lambda \approx 4.52 \times 10^{-12} \, \text{m} \]