Enter your answer in the provided box. what is the value of ni for an electron that emits a photon of wavelength 95.04 nm when it returns to the ground state in the h atom?

Ni = 5 The Rydberg formula for hydrogen is 1/w = R(1/a^2 - 1/b^2) where w = wavelength in vacuum R = Rydberg constant 1.0973731568508x10^7 1/m a,b = integers greater than or equal to 1 and a < b Now we need to select the value for a. a = 1 will converge towards 91.13 nm a = 2 converges towards 364.51 nm a = 3 converges towards 820.14 nm ... Because of this, we will assume a = 1 for this problem since it converges closest to the wavelength given. Substitute known values 1/w = R(1/a^2 - 1/b^2) 1/9.504x10^-8 = 1.0973731568508x10^7(1/1^2 - 1/b^2) 10521885.52 = 1.0973731568508x10^7(1/1 - 1/b^2) 0.958824759 = 1 - 1/b^2 -0.041175241 = -1/b^2 0.041175241 = 1/b^2 24.28643927 = b^2 4.928127359 = b So Ni = 5.

Atransverse wave (lowest wave) and a the transverse  wave's wavelength .