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: `(h = 6.63 xx 10^(-34)Js)`

Options

A. `8.041 xx 10^(-40) J`
B. `2.680xx 10^(-19)J`
C. `1.640xx10^(-18) J`
D. `6.111 xx 10^(-17)J`

Explanation

To calculate the energy of a single photon emitted by an excited hydrogen atom, we can use the formula:

\[ E = h \cdot \nu \]

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

### Step-by-Step Solution:

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

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

**Step 3: Perform the multiplication.**
- First, multiply the coefficients:
\[ 6.63 \times 2.47 = 16.3731 \]
- Then, multiply the powers of ten:
\[ 10^{-34} \times 10^{15} = 10^{-19} \]
- Combine these results:
\[ E = 16.3731 \times 10^{-19} \, \text{J} \]

**Step 4: Express the final answer in scientific notation.**
\[ E \approx 1.63731 \times 10^{-18} \, \text{J} \]

### Final Answer:
The energy of a single photon emitted by the excited hydrogen atom is approximately \( 1.64 \times 10^{-18} \, \text{J} \) (rounded to three significant figures).

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