Question

The first energy level for an element lies at 1.154 x 10-20 J and the second energy level lies at 3.460 x10-201. Determine the wavelength and frequency of the photon emitted when an electron drops from the second energy level to the first level. In which region of the electromagnetic spectrum will I find that photon.

Answer 1

Given that energy of first energy level (E_{​​​​​​1}) = 1.154×10^-20 J

The energy of second energy level (E_{​​​​​​2}) = 3.460×10^-20 J

When electron drops from 2nd energy level to first energy level, the energy of emitted photon is

∆E = E_{​​​​​​2} - E_{​​​​​​1}

∆E = 3.460×10^-20 J -1.154×10^-20 J

= 2.306×10^-20 J

Wavelength and energy of photon are related as

∆E = hc/λ

λ = hc/ ∆E

Where h is a Planck's constant = 6.625×10^-34 J sec

C is the velocity of light = 3×10⁸ m/sec

λ = 6.625×10^-34 J sec * 3×10⁸ m/sec /2.306×10^-20 J

= 8.6188×10^-6

  = 8618.8×10^-9

= 8618.8 nm

Energy and frequency are related as

∆E = hν

ν = ∆E/h

= 2.306×10^-20 J /6.625×10^-34 J sec

= 0.348 ×10¹⁴ Hz

= 3.48×10¹³ Hz

This energy region corresponds to Infrared region.