Question

The following is an energy-level diagram illustrating three different electronic transitions in the Bohr hydrogen atom.

a. Explain why the energy levels get closer together as they increase. Provide mathematical support for this.

b. Verify that the colours given in the diagram are correct. Provide mathematical support.

Short answer

Answer

The wavelength for red, green and blue light also correspond to wavelength values of 657. 1nm , 486. 8nm and 165. 28nm therefore the coloured lines depicted are correct.

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:

$En=\left(-RH\right)Zn2$

Where,

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

Z represent an atomic number.

n denotes a particular stationary state.

En denotes energy associated with a particular stationary state

Hydrogen emission spectrum that shows characteristics lines could be accounted for by the assumption made in Bohr's model. Since 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:

$∆E=h.∆v$

Where,

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

$∆E$ denotes energy absorbed or emitted in transition.

 Explain the energy levels get closer together as they increase. 

a)

The quantized energy for each of such discrete orbit can be obtained from the expression as follows:

$En=\left(-RH\right)Z2n2$

This expression suggests En is inversely related to the square of n . So higher the value of a particular stationary state more is the energy associated with it. Hence for consecutive energy levels, the energy difference is almost minimal. So, they begin to appear as a continuum.

Given information from the question

b)

The quantized energy for each of such discrete orbit can be obtained from the expression as follows:

$En=\left(-RH\right)Z2n2$ …….. (1)

For n=3

The value of RH is $2.178\times {10}^{-18}J$.

The value of Z for hydrogen atom is 1.

Substitute the values in equation (1),

$$\begin{gathered} E3=\left(-RH\right)Z2n2 \\ =\left(-2.178\times {10}^{18}J\right)1\left(3\right)2 \end{gathered}$$

For n=2

Substitute the values in equation (2),

$E2=\left(-RH\right)Z2n2$

$=\left(-2.178\times {10}^{-18}J\right)1\left(2\right)2$

The difference in these energies gives energy for red colour transition is calculated as follows:

$$\begin{gathered} E3-E2=\left(-2.178\times {10}^{-18}J\right)\left[1\left(3\right)2-1\left(2\right)2\right] \\ =3.025\times {10}^{-19}J \end{gathered}$$

Similarly, energy for green colour transitions as follows:

$$\begin{gathered} E4-E2=\left(-2.178\times {10}^{-18}J\right)\left[1\left(4\right)2-1\left(2\right)2\right] \\ =4.083\times {10}^{-19}J \end{gathered}$$

Likewise, energy for blue colour transitions as follows:

$$\begin{gathered} E5-E2=\left(-2.178\times {10}^{-18}J\right)\left[1\left(5\right)2-1\left(2\right)2\right] \\ =5.445\times {10}^{-19}J \end{gathered}$$

Formula to compute wavelength from energy difference is given as follow:

$\lambda =h.c∆E$ ……. (2)

Value of $∆E$ is $3.025\times {10}^{-19}J$.

Value of h is $6.626\times \times {10}^{-34}Js$.

Value of c is $3\times {10}^{8}m/s$.

Substitute the value in equation (2) to calculate the wavelength for the red colour.

$$\begin{gathered} \lambda =h.c∆E \\ =\left(6.626\times {10}^{-34}Js\right)\left(3\times {10}^{8}m/s\right)\left(3.025\times {10}^{19}J\right) \\ =\left(6.571\times {10}^{7}m\right)\left(1nm{10}^{-9}m\right) \\ =657.1nm \end{gathered}$$

The corresponding wavelength is calculated as follows;

Value of $∆E$ is $4.083\times {10}^{-19}J$.

Substitute the value in equation (2) to calculate the wavelength for the green colour.

$$\begin{gathered} \lambda =h.c∆E \\ =\left(6.626\times {10}^{-34}Js\right)\left(3\times {10}^{8}m/s\right)\left(4.083\times {10}^{-19}J\right) \\ =\left(4.868\times {10}^{-7}m\right)\left(1nm{10}^{-9}m\right) \\ =486.8nm \end{gathered}$$

Value of $∆E$ is $5.445\times {10}^{-19}J$.

Substitute the value in equation (2) to calculate the wavelength for the blue colour.

$$\begin{gathered} \lambda =h.c∆E \\ =\left(6.626\times {10}^{-34}Js\right)\left(3\times {10}^{8}m/s\right)\left(5.445\times {10}^{-19}J\right) \\ =\left(1.6528\times {10}^{-7}m\right)\left(1nm{10}^{-9}m\right) \\ =165.28nm \end{gathered}$$

Since the wavelength for red, green and blue light also correspond to wavelength values of 657. 1nm, 486. 8nm and 165. 28nm therefore the coloured lines depicted are correct.