Question

Consider a sample of excited atoms that lie 3.077 × 10–19 J above the ground state. Determine the emission wavelength (in nanometers) of these atoms.

Answer 1

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Energy = hv (h = Planck Constant and v = frequency = C/lambda, C = Velocity of light)

3.404 x 10^-19 = hC/lambda = 6.626 x 10^-34 x 3 x 10^8/lambda

wave length or lambda= 6.626 x 10^-34 x 3 x 10^8/3.404 x 10^-19 = 5.8396 x 10^-7 m = 583.96 nm

Answer 2

by using the equation
Energy = hv (h = Planck Constant and v = frequency = C/lambda, C = Velocity of light)
3.404 x 10^-19 = hC/lambda = 6.626 x 10^-34 x 3 x 10^8/lambda
wave length or lambda= 6.626 x 10^-34 x 3 x 10^8/3.404 x 10^-19 = 5.8396 x 10^-7 m = 583.96 nm
hope it helps