The answer is simple. Input Power = Output Power + Losses,.
One can overthink a problem. At the macro Newtonian scale Laws Of Thermodynamics rule.
Energy, heat, work, mass, conservation of energy and mass, and no lossless systems.
Without knowing anything about fluid mechanics, turbines, and electrical generators.
Required input power is output power/efficiency.
Water power in = 100 Megawatts/0.9 = 111 Megawatts
Making a broad assumption that the bulk of losses show up as heat the heat generated is 11 megawatts.
Find time to free fall 200 meters
h = .5*g*t^2
t = sqrt(200/(.5*9.8) = 6.3887656 seconds
Find velocity
v = g*t
v = 6.3887656 * 9.8 = 62.609903 meters/second
Find mass of water required
kinetic energy = .5*m*v^2
m = 111megawatts/(.5* 62.609903^2) = 5439000 kilograms
Find volume flow rate
At standard temperature and pressure density of water is approximately 1000kg/m^3
volume flow rate = 5439000/1000 = 5439 m^3/s
The discharge rate at the mouth of the Columbia is 7550 m^3/s
The Grand Coulee dam on then Columbia has a capacity of 6800 megawatts and a height of 168 meters.
Without a grounding in Newtonian mechanics and thermodynamic jumping to quantum and relativistic mechanics can lead to misinterpretations.
LOT does apply but the formulation changes.
In a photo detector quantum efficiency is photons in versus electrons out. At the quantum scale the efficiency of a photo detector can not be >= 1 as in any Newtonian scale system like a power pant.
For a photo voltaic solar cell it is( incident photons in watts) X (efficiency) = (outpit in watts), terminal voltage times current.
Planck’s constant expresses a photon’s energy which takes into account frequency..
h = 6.62607015×10−34 J⋅Hz−1 energy in joule
In solid state photo detectors the energy of photon and the band gap energy of the material are expressed in electron volts.
An electron volt (eV) is a unit of energy, specifically the amount of kinetic energy gained by a single electron when it accelerates through an electric potential difference of one volt. It's a very small unit, commonly used in atomic and nuclear physics.
One eV is equal to 1.602 x 10⁻¹⁹ joule
Energy always reduces to Joules. eV is a different way of defining energy, IOW the capacity to do work.
In photo detectors it is said when the energy of the photon equals the energy of the material band gap absorption occurs, both in eV. Not the same context as saying coal has energy in J/kg, but energy non the less..