Question

A metal ion Mⁿ⁺ has a single electron. The highest energy line in its emission spectrum occurs at a frequency of 2.961 x 10¹⁶ Hz. Identify the ion.

Answer 1

Answer:

z≅3

Atomic number is 3, So ion is Lithium ion ($Li^+$)

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

$\lambda=\frac{3*10^8}{2.961*10^{16}}\\ \lambda=1.01317*10^{-8} m$

Using Rydberg's Formula:

$\frac{1}{\lambda}=R*z^2*(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

Where:

R is Rydberg constant=$1.097*10^7$

z is atomic Number

For highest Energy:

n_1=1

n_2=∞

$\frac{1}{1.01317*10^{-8}}=1.097*10^{7}*z^2*(\frac{1}{1^2}-\frac{1}{\inf})\\z^2=8.99\\z=2.99$

z≅3

Atomic number is 3, So ion is Lithium ion ($Li^+$)