Question

Energy required for the electron excitation of $L{i^{ + + }}$ from the first to the third Bohr state orbit is:
A) $12.1eV$
B) $36.3eV$
C) $108.8eV$
D) $122.4eV$

Answer 1

We know that Li has an atomic number(Z) of 3, It means it has 3 electrons and we are given $L{i^{ + + }}$ state of $Li$. For energy required to change the state we first need to find the energy of the lower state (first state) then find the energy at the higher state (third state) then take the difference of them the value obtained is the energy required for the state transfer of $L{i^{ + + }}$ ion.

Complete step by step answer:
Energy of an electron in $n^{th}$ orbit can be given as ${E_n} = \dfrac{{ - 13.6{Z^2}}}{{{n^2}}}$.
Now for $L{i^{ + + }}$ ion, Z=3
And in lower state, n=1
In third state (higher state), n=3
Using the given above formula we calculate energy in each state as:
For first or lower state energy is,
${E_1} = \dfrac{{ - 13.6({3^2})}}{{{1^2}}}$
For higher or third state energy is,
${E_3} = \dfrac{{ - 13.6({3^2})}}{{{3^2}}}$
Now we have got the energy at two different state so now let’s take their difference,
$\Delta E = E2 - E1$
$\Delta E = - 13.6 - ( - 122.4) = 108.8\;eV$

Therefore, the energy required for state transfer is: $108.8\;eV$. Hence, option C is correct.

Note:
In Bohr’s model of an atom, Here, $\Delta E$ is the change in energy between the initial and final orbits, and $hv$ is the energy of the absorbed or emitted photon. It is quite logical that energy is involved in changing orbits. A blast of energy is required for the space shuttle, for example, to climb to a higher orbit. What is not expected is that atomic orbits should be quantized. This is not observed for satellites or planets