Excited hydrogen atom emits light in the ultraviolet region at `2.47xx 10^(15)` Hz. With this frequency, the energy of a single photon is:

Options

A. `8.041 xx 10^(-40) j`
B. `2.680 xx 10^(-19) J`
C. `1.640 xx 10^(-18)J`
D. `6.111 xx 10^(-17)J`

Explanation

To find the energy of a single photon emitted by an excited hydrogen atom at a frequency of \(2.47 \times 10^{15}\) Hz, we can use the formula:

\[ E = h \nu \]

where:
- \(E\) is the energy of the photon,
- \(h\) is Planck's constant, approximately \(6.63 \times 10^{-34} \, \text{J s}\),
- \(\nu\) is the frequency of the light, which is given as \(2.47 \times 10^{15} \, \text{Hz}\).

### Step-by-Step Solution:

1. **Identify the given values:**
- Frequency (\(\nu\)) = \(2.47 \times 10^{15} \, \text{Hz}\)
- Planck's constant (\(h\)) = \(6.63 \times 10^{-34} \, \text{J s}\)

2. **Substitute the values into the formula:**
\[ E = (6.63 \times 10^{-34} \, \text{J s}) \times (2.47 \times 10^{15} \, \text{Hz}) \]

3. **Perform the multiplication:**
- First, multiply the numerical values:
\[ 6.63 \times 2.47 \approx 16.36 \]
- Next, multiply the powers of ten:
\[ 10^{-34} \times 10^{15} = 10^{-19} \]

4. **Combine the results:**
\[ E \approx 16.36 \times 10^{-19} \, \text{J} \]

5. **Express in scientific notation:**
\[ E \approx 1.636 \times 10^{-18} \, \text{J} \]

6. **Round to two decimal places:**
\[ E \approx 1.64 \times 10^{-18} \, \text{J} \]

### Final Answer:
The energy of a single photon emitted by the excited hydrogen atom is approximately:
\[ E \approx 1.64 \times 10^{-18} \, \text{J} \]