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 during a transition between two energy levels in an atom, we can use the relationship between energy and frequency, which is derived from Planck's equation. This relationship is expressed as:

Understanding the Relationship

The equation we need is:

E = h * f

Where:

• E is the energy difference between the two levels (in joules).
• h is Planck's constant, approximately 6.626 x 10^-34 J·s.
• f is the frequency of the emitted radiation (in hertz).

Converting Energy Units

In your case, the energy difference is given as 2.3 eV. To use this in our equation, we need to convert electronvolts to joules. The conversion factor is:

1 eV = 1.602 x 10^-19 J

So, to convert 2.3 eV to joules:

E = 2.3 eV * 1.602 x 10^-19 J/eV = 3.6886 x 10^-19 J

Calculating Frequency

Now that we have the energy in joules, we can rearrange the equation to solve for frequency:

f = E / h

Substituting the values we have:

f = (3.6886 x 10^-19 J) / (6.626 x 10^-34 J·s)

Calculating this gives:

f ≈ 5.56 x 10^14 Hz

Final Thoughts

The frequency of the radiation emitted when the atom transitions from the upper energy level to the lower one is approximately 5.56 x 10^14 Hz. This frequency falls within the visible light spectrum, specifically in the range of blue to violet light, which is fascinating as it connects atomic transitions to observable phenomena in our everyday world.