Asked by Swapnil
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.
Answers
Answered by
DrBob222
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.
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.
Answered by
Amit Kumar Garanayak
6 to 1
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