Question

Using the Bohr model, determine the energy in joules of the photon produced when an electron in a Li2+ ion moves from the orbit with n = 2 to the orbit with n = 1

Answer 1

Bohr’s model of atom postulated that the electrons revolve around the nucleus only in those orbits which have a fixed amount of energy and do not lose energy while revolving in them and energy is gained or lost only when electrons move from one orbit to another.

According to Bohr’s model, the energy at infinite distance is taken to be zero and as it approaches the atom, it starts becoming more negative.

The difference in energy of an electron when it moves from one orbit to another in the hydrogen atom is calculated by using the following formula.

$$ \begin{aligned} \Delta \mathrm{E} &=-Z^{2} R_{H}\left(\frac{1}{n_{2}^{2}}-\frac{1}{n_{1}^{2}}\right)  \end{aligned} $$

Here ΔE is the difference in energy, Z is the atomic number of the atom, R_H is the Rydberg’s constant whose value is equal to  2.179×10^-18J, n₁ is the initial orbit of the electron and  n₂ is the final orbit of the electron.

 Li²⁺ is a hydrogen-like ion. Thus, the energy of the photon produced during the movement of an electron from n = 2 to n = 1 is calculated as follows;

$$ \begin{aligned} \Delta \mathrm{E} &=-Z^{2} R_{H}\left(\frac{1}{n_{2}^{2}}-\frac{1}{n_{1}^{2}}\right) \\ &=-(3)^{2}\left(2.179 \times 10^{-18}\right)\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right) \mathrm{J} \\ &=-1.47 \times 10^{-17} \mathrm{~J} \end{aligned} $$

The energy of the photon, when an electron in a Li²⁺ ion moves from the orbit with n = 2 to the orbit with n = 1 is  -1.47 × 10^-17J

Since the photon is moving down in energy levels the difference in energy (? E) is negative and the radiation is emitted.