the wave number of the limiting line in Balmer series of He+ would be...
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6
hello dear here is ur answer
hope it helps u dear
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but here ans.is 109678 cm inverse..can u give a calculated answer...
i need calculation of answer..please help me
ohh srry i calculated the answer of hydrogen atom
so please can you do for hellium one..please dear..
ok dear i will
but where to answer
just comment me here
but i have solved in it in a book so i cant post it in a comment
ok i will post it in ur any question
Answered by
2
Its very simple..
Use this formula, Wave number = R [(1/n1"2) - (1/n2"2)] *
772
Here, nl is the orbit corresponding to the series. In case
of our question, Balmer series corresponds to 2.
n2 is any other higher orbit. In the present case, the
limiting line is from infinity (ee).
As the atom is a hydrogen atom, Z=1.
R is the Rydberg Constant, R=109677 cm-1.
Now substitute all the values and you will get an
approximate answer as, 27419 cm-1.
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