A

To find the frequency of radiation emitted when an atom transitions between energy levels, we can use the formula that relates the energy difference between the levels (in electron volts, eV) to the frequency of the emitted radiation. The formula is given by:
$$ E = h u $$
where:
- $E$ is the energy difference (2.3 eV),
- $h$ is Planck's constant ($6.626 \times 10^{-34} \, \text{Js}$) and
- $\nu$ is the frequency in hertz (Hz).
To convert energy from eV to joules, we use the conversion factor:
$$ 1 \, eV = 1.602 \times 10^{-19} \, J $$
Therefore, the energy in joules is:
$$ E = 2.3 imes 1.602 \times 10^{-19} \, J = 3.6846 \times 10^{-19} \, J $$
Now, substituting the values into the formula:
$$ \nu = \frac{E}{h} = \frac{3.6846 \times 10^{-19} \, J}{6.626 \times 10^{-34} \, Js} $$
Calculating this gives:
$$ \nu \approx 5.56 \times 10^{14} \, Hz $$
Therefore, the frequency of the radiation emitted is approximately $5.56 \times 10^{14}$ Hz.