Q8 / 160 · AP EAMCET 2020 (21 Sep Shift 2) · Chemistry
Calculate energy of half mole of photons of a radiation with frequency \(3 \times 10^{12} \mathrm{~Hz}\).
- A\(598.2 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
- Correct answerB\(0.598 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
- C\(1.196 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
- D\(119.6 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
Solution
Energy of a photon, \(E=h v\), where, \(v=\) frequency \(=3 \times 10^{12} \mathrm{~Hz}\)
Therefore, energy
\(=6.6 \times 10^{-34} \times 3 \times 10^{12} \mathrm{~Hz}=1.9878 \times 10^{-21} \mathrm{~J}\)
1 mole \(=6.02 \times 10^{23}\) photons,
therefore, \(\frac{1}{2}\) mole \(=3.01 \times 10^{23}\) photons
Therefore, energy of \(\frac{1}{2}\) mole of photons
\(\begin{aligned}
& =1.9878 \times 10^{-21} \times 3.01 \times 10^{23} \\
& =0.5982 \mathrm{~kJ} / \mathrm{mol}
\end{aligned}\)
Hence, the correct option is (b).
Hence, the correct answer is option: 'B'.