Asked by Anisha
Consider a sample of excited atoms that lie 3.211 × 10–19 J above the ground state. Determine the emission wavelength (in nanometers) of these atoms.
Answers
Answered by
Damon
E = h f
3.2 * 10^-19 = 6.6*10^-34 f
f = .485 * 10^15 Hz
T = 1/f = 2.06 * 10-15 seconds
if c = 3*10^8 m/s
lambda =c T =3*10^8 * 2.06*10^-15
= 6.19 * 10^-7 meters
= 619 * 10^-9 meters
3.2 * 10^-19 = 6.6*10^-34 f
f = .485 * 10^15 Hz
T = 1/f = 2.06 * 10-15 seconds
if c = 3*10^8 m/s
lambda =c T =3*10^8 * 2.06*10^-15
= 6.19 * 10^-7 meters
= 619 * 10^-9 meters
Answered by
Jose B
((6.626*10^-34)(3.00*10^8))/(3.332*10^-19)
= 5.966*10^-7
in nanometers it is 596.6 nm
= 5.966*10^-7
in nanometers it is 596.6 nm
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