Question

whar transition in He^+ ion shall have the same wave number as the first line in Balmer series of hydrogen atom​

Answer 1

answer : n1 = 4 and n2 = 6

formula of wave number is given by,

$\frac{1}{\lambda}=RZ^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

where R is Rydberg's constant, Z is atomic number, $n_1,n_2$ are transition of electron from first to second respectively.

for first balmer series,

n1 = 2 and n2 = 3

and Z of hydrogen atom is 1.

so, wave number $\frac{1}{\lambda}$ = R(1)² [ 1/2² - 1/3²] = 5R/36....(1)

now for He^+ , Z = 2

and then wave number , $\frac{1}{\lambda}$ = R(2)² [ 1/n1² - 1/n2²] ....(2)

a/c to question,

wave number of He^+ = wave number of first line in balmer series of H - atom

or, 5R/36 = 4R[1/n1² - 1/n2² ]

or, 5/144 = [1/n1² - 1/n2²]

or, (1/4² - 1/6² ) = [ 1/n1² - 1/n2²]

on comparing both sides,

n1 = 4 and n2 = 6