The Rydberg equation for the hydrogen atom is given by (1/λ) = RH (1/nf^2 - 1/ni^2)
Where RH is the hydrogenic Rydberg constant, 1.0974×10^7m^−1, nf and ni are the final and initial principal quantum numbers of the energy levels, respectively. (The transition is from ni to nf )
The Balmer series are emission lines associated with transitions ending at the n = 2 principal level.
What is the upper principal level that generates a transition at 656.3 nm?
(1nm=10^−9m)

Answer: a charge of -1.0  × 10⁻⁴ c must be placed  at x = 26 m.the sign of the charge will be the same sign as the  -4.0  µc charge.======electricstatic force between two point charges  is given by    where k is coulomb's constant,  8.99×10⁹  n·m²/c², q₁ and q₂ are the point charges, and r is the distance between the two point charges.assign positive to q₁let q₁ =  6.0 µc, q₂ = -4.0  µc, q₀ represent the charge at the origin, and q₃ represent this new charge to place.the charge at the origin q₀ would experience the following force f₀₁ from q₁, with r = 6.0 m (note that 1  µc = 10⁻⁶ c)    the force from q₂    the force from q₃    the net force on the origin charge, q₀, has to be zero for no electrostatic forcehaving an equation where k and q₀ can be divided out of the equation will allow us to solve for q₃    isolating:     a charge of -1.0  × 10⁻⁴ c must be placed  at x = 26 m.the sign of the charge will be the same as the  -4.0  µc charge