Two of the emission wavelengths in the hydrogen emission spectrum are 656 nm and 486 nm. One of these is due to the n = 4 to n = 2 transition, and the other is due to the n = 3 to n = 2 transition. Which wavelength goes with which transition?

e n = 3 to n = 2 =  486 nm

                n = 4 to n = 2 = 656 nm

Explanation: Solution

E = hf = hc/λ.  

Here h = 6.626*10-34J/S  ,  

C= Speed of ligfht=3*10^8m/s and

λ = wavelength

Since h and c are constants, we now have  that the energy of photon is inversely proportional to the wavelengths of the speed of light .  

E ᵅ 1/ λ  ie higher energy given lower wavelength

In this question, we were given  wavelengths of 656 nm and 486 nm

The principle is that The transitions between orbit that are closer together produce lower energy lights and therefore higher wavelengths  than  orbits that are further apart.

Therefore For transition  due to the n = 3 to n = 2, it will have wavelength number of 486 nm since orbits are closer.

For transition  due to the n = 4 to n = 2, it will have wavelength number of  656 nm since orbits are farther apart

2b2t

explanation:

2b2t