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vova2212 [387]
3 years ago
11

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.
Chemistry
1 answer:
Whitepunk [10]3 years ago
3 0

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^+)

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