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.

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

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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 

The activity series is shown in the attached picture. if the element is below the other, then you can displace the other. following the activity series as a guide, the reactions are as follows: nickel and aluminum chloride3 ni + 2 alcl₃  → 3 nicl₂ + 2 albarium and copper bromideba + cubr  →  no reaction because cu is less oxidizable than bapotassium nitrate and aluminum3 kno₃ + al  → al(no₃)₃ + 3 ksodium phosphate and zincnaso₄ + zn  → znso₄ + na