Which one of the following electronic transitions in hydrogen atom will lead to the emission of the SHORTEST wavelength, n=1 to n=4, n=4 to n=2, n=3 to n=2?Justify with reason!​

The shortest wavelength corresponds to the highest energy transition. In a hydrogen atom, the energy difference between levels increases as you get closer to the nucleus. Therefore, the transition from n=4 to n=1 will have the highest energy and thus the shortest wavelength.

Here's why:

The energy of an electron in a hydrogen atom is given by the formula E = -13.6/n^2 eV, where n is the principal quantum number. The energy difference (and therefore the energy of the photon emitted in the transition) is given by the absolute difference in energy between the two levels.

For n=1 to n=4, the energy difference is |-13.6/1^2 - (-13.6/4^2)| = 12.75 eV. For n=4 to n=2, the energy difference is |-13.6/4^2 - (-13.6/2^2)| = 3.40 eV. For n=3 to n=2, the energy difference is |-13.6/3^2 - (-13.6/2^2)| = 1.89 eV.

So, the transition from n=4 to n=1 has the highest energy and will therefore have the shortest wavelength, because energy and wavelength are inversely proportional (E = hc/λ, where h is Planck's constant, c is the speed of light, and λ is the wavelength).