A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom make a transition from the upper level to the lower level?
The frequency of radiation emitted is approximately ¹⁴ Hz.
Explanation
According to Bohr's model, when an atom makes a transition from a higher energy state to a lower energy state, the energy difference is carried away by a photon. The frequency of this photon is given by the relation hν = E_i - E_f (Equation 12.11 in the textbook). The energy difference must be converted from eV to Joules before calculating the frequency using Planck's constant.
Solution Steps
Step 1: Identify the formula from Bohr's model: hν = E_i - E_f, where ΔE = 2.3 eV
Step 2: Convert energy from eV to Joules: ΔE = × 10⁻¹⁹ J = ⁻¹⁹ J
Step 3: Calculate frequency using h = ⁻³⁴ J·s: ν = ΔE/h = (⁻¹⁹)/(⁻³⁴) = ¹⁴ Hz