Excited hydrogen atom emits light in the ultraviolet region at $$2.47 \times 10^{15}$$ Hz. With this frequency, the energy of a single photon is: (h = $$6.63 \times 10^{-34}$$ Js)

The energy of a single photon is calculated using the formula $$E = h\nu$$, where $$h$$ is Planck's constant and $$\nu$$ is the frequency of the light.

Given:

• Frequency, $$\nu = 2.47 \times 10^{15}$$ Hz
• Planck's constant, $$h = 6.63 \times 10^{-34}$$ J s

Substitute the values into the formula: $$$ E = (6.63 \times 10^{-34}) \times (2.47 \times 10^{15}) $$$

First, multiply the coefficients: $$$ 6.63 \times 2.47 $$$

Break it down step by step: $$$ 6.63 \times 2.47 = 6.63 \times (2 + 0.4 + 0.07) $$$ $$$ = (6.63 \times 2) + (6.63 \times 0.4) + (6.63 \times 0.07) $$$ $$$ = 13.26 + 2.652 + 0.4641 $$$ $$$ = 16.3761 $$$

Now, multiply the powers of 10: $$$ 10^{-34} \times 10^{15} = 10^{-34 + 15} = 10^{-19} $$$

Combine the results: $$$ E = 16.3761 \times 10^{-19} \text{ J} $$$

Convert to standard scientific notation by adjusting the decimal: $$$ 16.3761 \times 10^{-19} = 1.63761 \times 10^1 \times 10^{-19} = 1.63761 \times 10^{-18} \text{ J} $$$

Consider significant figures. Both given values ($$6.63$$ and $$2.47$$) have three significant figures. Round $$1.63761 \times 10^{-18}$$ to three significant figures:

• The third digit is $$3$$ (in $$1.63$$), and the next digit $$7 \geq 5$$, so round up: $$1.64 \times 10^{-18}$$ J.

Compare with the options:

• A: $$8.041 \times 10^{-40}$$ J
• B: $$2.680 \times 10^{-19}$$ J
• C: $$1.640 \times 10^{-18}$$ J
• D: $$6.111 \times 10^{-17}$$ J

Hence, the correct answer is Option C.