A sample contains Hydrogen atom ,He+ ion, Li2+, & Be3+ ion. In H atom electron is present in 8th orbit, in He2+ e- is present in 6th orbit, Li2+ in 5th orbit& in Be3+ electron is present in 4th orbit. All the atoms are de-excited to the ground state. Calculate the total no. of different spectral lines.
6 to 1
To calculate the total number of different spectral lines, we need to consider the transitions that can occur between the energy levels of the atoms.
The number of spectral lines produced when an electron transitions from one energy level to another can be given by the formula:
N = n(n-1)/2,
where 'n' is the number of energy levels.
In the case of the Hydrogen atom, the electron is present in the 8th orbit and is de-excited to the ground state, which is the first orbit (n = 1). So, the number of spectral lines produced by the Hydrogen atom would be:
N_hydrogen = 8(8-1)/2 = 8*7/2 = 28.
For the He2+ ion, the electron is present in the 6th orbit and is de-excited to the ground state (n = 1). Thus, the number of spectral lines produced by the He2+ ion is:
N_helium = 6(6-1)/2 = 6*5/2 = 15.
Similarly, for the Li2+ ion with the electron in the 5th orbit (n = 1) and the Be3+ ion with the electron in the 4th orbit (n = 1), the number of spectral lines produced would be:
N_lithium = 5(5-1)/2 = 5*4/2 = 10.
N_beryllium = 4(4-1)/2 = 4*3/2 = 6.
Therefore, the total number of different spectral lines produced by the given atoms would be the sum of the spectral lines calculated for each atom:
Total spectral lines = N_hydrogen + N_helium + N_lithium + N_beryllium
= 28 + 15 + 10 + 6
= 59.
Hence, there would be a total of 59 different spectral lines.
For H^+, e is in n = 8 so the follwing transitions can occur.
n = 8 to n = 7
8 to 6
8 to 5
8 to 4
8 to 3
8 to 2
and 8 to 1;
then
n = 7 to n = 6
7 to 5
7 to 4
7 to 3
7 to 2
and 7 to 1; then
n = 6 to n = 5
6 to 4
6 to 3
6 to 2
and 6 to 1
You get the idea here. Do the same for the other ions and add for the total number of lines.