Question

Consider only the transitions involving the first four energy levels for a hydrogen atom:

a. How many emissions are possible for an electron in the n=4level as it goes to the ground state?

b. Which electronic transition is the lowest energy?

c. Which electronic transition corresponds to the shortest wavelength emission?

Short answer

a) To arrive at the ground state, three transitions are possible.

b) Transition from the second energy level to the ground state is of the lowest energy.

c) The states involved in such transition are$\mathrm{n}$ equal to 4 and$\mathrm{n}$ equal to 1 for first four energy states in an$\mathrm{H}$ atom.

Step-by-step solution

Concept used

Salient characteristics of Bohr's model are as follows:

Electrons revolved in certain discrete circular orbits that were associated with fixed angular momentum.

Such discrete orbits are termed non-radiating or stationary orbits. The quantized energy for each of such discrete orbit can be obtained from the expression as follows:

$\mathrm{En}=(-\mathrm{RH})\mathrm{Z}2\mathrm{n}2$

Where,

$\mathrm{RH}$is constant that has a value of$2.178\times {10}^{-18}\mathrm{J}$

$\mathrm{Z}$represent an atomic number.

$\mathrm{n}$denotes a particular stationary state.

$\mathrm{En}$denotes energy associated with a particular stationary state

Calculate the possible transition

a)

Energy absorption or emission only occurred when an electron jumped from one energy state to another hence each such probable unique transition gave rise to specific lines in the emission spectrum of the hydrogen atom. This was based on the relationship as follows:

$∆\mathrm{E}=\mathrm{h}.∆\mathrm{V}$

Where,

$∆\mathrm{v}$is a difference in the frequency of two stationary states involved in the transition.

$∆\mathrm{E}$denotes energy absorbed or emitted in transition.

For an electron situated in$\mathrm{n}$ equal to$4$ the probable transition occurs to$\mathrm{n}$ equal to $3$, $\mathrm{n}$equal to $2$and$\mathrm{n}$ equal to $1$. Thus, to arrive at the ground state, three transitions are possible.

Calculate the lowest energy transition

b)

The expression of quantized energy for each discrete orbit is as follows:

$∆\mathrm{E}=\mathrm{h}.\mathrm{c\lambda }$

Since lowest-energy corresponds to maximum wavelength or least frequency the states involved in such transition $\mathrm{n}$are equal to 2 and$\mathrm{n}$ equal to 1. Thus, the transition from the second energy level to the ground state is of the lowest energy.

Identify the transition that corresponds to the shortest wavelength

c)

The expression of quantized energy for each discrete orbit is as follows:

$∆\mathrm{E}=\mathrm{h}.\mathrm{c\lambda }$

In order to have the shortest wavelength, the frequency must be highest so the energy difference should be highest. Therefore, the states involved in such transition are$\mathrm{n}$ equal to$4$ and$\mathrm{n}$ equal to$1$ for first four energy states in an$\mathrm{H}$ atom.