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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 th

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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.

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