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?

To find the frequency of radiation emitted when an atom makes a transition between two energy levels, we use the equation derived from Planck’s relation:

Energy of a photon:

E = hν

where,
E = energy difference between levels (in joules),
h = Planck’s constant (6.626 × 10⁻³⁴ J·s),
ν = frequency of emitted radiation (in Hz).

Step 1: Convert energy difference from eV to Joules
Given energy difference:

E = 2.3 eV

We use the conversion factor:

1 eV = 1.602 × 10⁻¹⁹ J

Thus,

E = 2.3 × (1.602 × 10⁻¹⁹) J
= 3.6846 × 10⁻¹⁹ J

Step 2: Use Planck’s equation to find frequency
E = hν

Rearranging for ν:

ν = E / h

Substituting values:

ν = (3.6846 × 10⁻¹⁹ J) / (6.626 × 10⁻³⁴ J·s)

ν = 5.56 × 10¹⁴ Hz

Final Answer:
The frequency of radiation emitted is 5.56 × 10¹⁴ Hz.