Answer:
z≅3
Atomic number is 3, So ion is Lithium ion ([tex]Li^+[/tex])
Explanation:
First of all
v=f*λ
In our case v=c
c=f*λ
λ=c/f
where:
c is the speed of light
f is the frequency
[tex]\lambda=\frac{3*10^8}{2.961*10^{16}}\\ \lambda=1.01317*10^{-8} m[/tex]
Using Rydberg's Formula:
[tex]\frac{1}{\lambda}=R*z^2*(\frac{1}{n_1^2}-\frac{1}{n_2^2})[/tex]
Where:
R is Rydberg constant=[tex]1.097*10^7[/tex]
z is atomic Number
For highest Energy:
n_1=1
n_2=∞
[tex]\frac{1}{1.01317*10^{-8}}=1.097*10^{7}*z^2*(\frac{1}{1^2}-\frac{1}{\inf})\\z^2=8.99\\z=2.99[/tex]
z≅3
Atomic number is 3, So ion is Lithium ion ([tex]Li^+[/tex])
