Question

thinking critically
answer the following questions.

1. a radio station has a frequency of 103.7 mhz. (1 mhz = 10^6 s^-1) what is the wavelength of the radiation emitted by the station? indicate where this wavelength falls on the electromagnetic spectrum shown below.

1. look at the electromagnetic spectrum again. are the microwaves used to cook food higher or lower in frequency than radio waves? are microwaves longer or shorter in wavelength than radio waves?

1. write the complete electron configuration and the noble - gas notation for aluminum.

complete
noble gas notation

1. write the orbital diagram of aluminum.

1. write the noble - gas notation of iodine.

1. identify each atom.

a. 1s^22s^22p^1
b. ar4s^1

1. give the number of valence electrons for the following atoms.

a. neon
c. carbon
b. hydrogen
d. sulfur

Explanation:

Question 1

Step1: Convert frequency to Hz

Given \(1\ MHz = 10^{6}\ s^{-1}\), so \(103.7\ MHz=103.7\times10^{6}\ Hz = 1.037\times 10^{8}\ Hz\)

Step2: Use the formula \(c = \lambda

u\) (where \(c = 3\times10^{8}\ m/s\) is the speed of light, \(\lambda\) is wavelength and \(
u\) is frequency)
Rearrange for \(\lambda\): \(\lambda=\frac{c}{
u}\)

Substitute \(c = 3\times 10^{8}\ m/s\) and \(
u=1.037\times 10^{8}\ Hz\)

\(\lambda=\frac{3\times 10^{8}}{1.037\times 10^{8}}\ m\approx 2.9\ m\)

From the electromagnetic spectrum, this wavelength falls in the FM radio range.

Looking at the electromagnetic spectrum:

• Frequency: Microwaves have a higher frequency than radio waves (since on the frequency axis \(v\) (Hz), microwaves are to the right of radio waves)
• Wavelength: Using \(c = \lambda

u\) (inverse relationship), since \(v_{microwave}>v_{radio}\), \(\lambda_{microwave}<\lambda_{radio}\) (microwaves are shorter in wavelength than radio waves)

• Complete electron configuration:

Aluminum (\(Al\), atomic number \(Z = 13\))
Filling orbitals in order \(1s<2s<2p<3s<3p\)
\(1s^{2}2s^{2}2p^{6}3s^{2}3p^{1}\)

• Noble - gas notation:

The noble gas before \(Al\) is neon (\(Ne\), \(1s^{2}2s^{2}2p^{6}\))
So \([Ne]3s^{2}3p^{1}\)

Answer:

The wavelength is approximately \(2.9\ m\) and it falls in the FM radio range of the electromagnetic spectrum.

Question 2