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. (a) `4.5xx10 ^(-7) m`
B. (b) ` 4.5 xx 10 ^(-7) nm`
C. (c) ` 4.5 xx 10 ^(-7) Å`
D. (d) `4.5 xx 10 ^(-7 ) cm`

Answer

To find the wavelength of the photon required to produce a transition from the ground state to the excited state of an atom, we can use the formula that relates energy difference (ΔE) to wavelength (λ):

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

Where:
- \( \Delta E \) is the energy difference (in joules)
- \( h \) is Planck's constant (\(6.63 \times 10^{-34} \, \text{J s}\))
- \( c \) is the speed of light (\(3 \times 10^{8} \, \text{m/s}\))
- \( \lambda \) is the wavelength (in meters)

### Step-by-Step Solution:

1. **Identify the given values**:
- Energy difference, \( \Delta E = 4.4 \times 10^{-19} \, \text{J} \)
- Planck's constant, \( h = 6.63 \times 10^{-34} \, \text{J s} \)
- Speed of light, \( c = 3 \times 10^{8} \, \text{m/s} \)

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

3. **Substitute the known values into the equation**:
\[ \lambda = \frac{(6.63 \times 10^{-34} \, \text{J s}) \times (3 \times 10^{8} \, \text{m/s})}{4.4 \times 10^{-19} \, \text{J}} \]

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

5. **Now divide by the energy difference**:
\[ \lambda = \frac{1.989 \times 10^{-25} \, \text{J m}}{4.4 \times 10^{-19} \, \text{J}} \]

6. **Perform the division**:
\[ \lambda = 4.52 \times 10^{-7} \, \text{m} \]

7. **Express in scientific notation**:
\[ \lambda \approx 4.5 \times 10^{-7} \, \text{m} \]

### Final Answer:
The wavelength of the photon required to produce this transition is approximately \( \lambda = 4.5 \times 10^{-7} \, \text{m} \).

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