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Furkat [3]
3 years ago
15

The ratio of the wavelength of AM radio waves traveling in a vacuum to the wavelength of FM radio waves traveling in a vacuum is

approximately
1.) 1 to 1
2.) 2 to 1
3.) 10^2 to 1
4.) 10^8 to 1
Physics
1 answer:
Debora [2.8K]3 years ago
3 0

This question is a big fat non sequitur !

The wavelength of radio waves traveling through vacuum only depends on the frequency that the radio station is licensed to broadcast on, (which had better be the frequency of the transmitter that they buy and use, or they're in big trouble).

The wavelength does NOT depend on the type of modulation that's used to put information onto the signal.  

An amateur radio (ham) operator may very well start out using FM to talk over his radio to somebody else, and then for some reason they may decide to switch to AM.  They can do that without ANY change in the wavelength of their transmissions.

Now, in the USA and many other countries, it so happens that all AM stations are licensed by their governments to transmit their programs on a channel somewhere between 500 KHz and 1.6 MHz, and all FM stations are licensed by their governments to transmit their programs on a channel somewhere between 88 MHz and 108 MHz.  (And THAT's what the radio receivers in these countries are built to receive.)

Then we might say that all of the AM stations are grouped around 1 MHz, and all of the FM stations are grouped around 100 MHz.  The FM frequencies are very roughly 100 times the AM frequencies, so the AM wavelengths are very roughly  100 times the FM wavelengths.  That's <em>choice (3)</em> .

But please don't get the idea that it has anything to do with using AM or FM technology.  It's just a matter of where in the spectrum the government decided to put the AM stations and where they put the FM stations.

For that matter . . . An analog TV station uses an AM signal for the picture and an FM signal for the sound, and it all goes in the same channel, with just about the same wavelengths !

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Answer:

B) Na has a lower first ionization energy than Ne.  

Explanation:

The atomic number¹ for Na has a value of 11 while in the case of Ne this value is 10. That means that Sodium (Na) has a total number of 11 protons, 11 neutrons and 11 electrons (since it is electrically neutral²). For the case of Neon (Ne) it has 10 protons, 10 neutrons and 10 electrons.

As the atomic number increases, the atomic radius³ shrinks (the orbitals are closer to the nucleus) as a consequence of the electric force. For the case of sodium (Na) the electron in the outermost orbital will experience a lower electric force than the electron placed in the outermost orbital in the atom of Neon (Ne).

Although, the sodium’s atom has more protons and therefore electrons, these eleven electrons will be organized according with the electronic configuration⁴ in the different shells (orbitals) of probabilities of their positions around the atom.

The electronic configuration for Na is:

1s²2s²2p⁶3s¹

The electronic configuration for Ne is:

1s²2s²2p⁶

Since Na needs another orbital to placed its outermost electron, the atomic radius will have a greater value than Ne. The electric force is inversely proportional to the square of the distance between two charged particles, as is established in Coulomb’s law:

F = \kappa_{0} \frac{q1q2}{r^{2}}    (1)

Where q1 and q2 are the charges, \kappa_{0} is the proportionality constant and r is the distance between the two charges.

Hence, the electron in the outermost orbital of Ne is submitted to a greater electric force according with equation 1, the required energy to remove it (ionization energy⁵) will be greater than in the case of Na (<u>for that case will be the first ionization energy</u>).

¹Atomic number: The number of protons or electrons in an atom.

²Electricaly neutral: All the charges are balanced (same number of positive charges and negative charges).

³Atomic radius: Distance between the center of the nucleus and an electron placed in the outermost orbital for a specific atom.

⁴Electronic configuration: Show how the electrons of an atom will be arranged in different orbitals according with the fact that each orbital has a specific number of electrons that can be held.

⁵Ionization energy: Energy required to remove an electron from an atom.

Key values:

First ionization energy of Na: 495 kJ/mol

First ionization energy of Ne: 2080 kJ/mol

Atomic radius of Na: 2.27 Å

Atomic radius of Ne: 1.54 Å

Atomic number of Na: 11

Atomic number of Ne: 10

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