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amid [387]
3 years ago
11

The atomic number of beryllium (Be) is 4, and the atomic number of barium (Ba) is 56. Which comparison is best supported by this

information?
Physics
2 answers:
Vesnalui [34]3 years ago
8 0
The atomic number of beryllium (Be) is 4, and the atomic number of barium (Ba) is 56. the <span>comparison is best supported by this information is that beryllium has a lower atomic radius than Barium</span>
nika2105 [10]3 years ago
8 0
<h3><u>Answer;</u></h3>

The comparison that would be best is;

<em>They are in the same group because they have similar chemical properties, but they are in different periods because they have very different atomic numbers.</em>

<h3><u>Explanation;</u></h3>
  • <em><u>Beryllium (Be) is an element is the periodic table with an atomic number of 4, which means it has 4 protons in the nucleus and also 4 electrons in the energy levels, since the number of protons is equal to the number of electrons.</u></em>
  • Beryllium has an electron configuration of 2: 2, which means it has two valence electrons, and has two energy levels which means<em><u> it occupies period two of the periodic table, and group two of the periodic table.</u></em>
  • Barium (Ba) on the other had, has an atomic number of 56, which means it has 56 protons and 56 electrons. Thus it has an electron configuration of; 2;8;18;18;8;2;  this shows that<u><em> it has 6 energy levels thus occupies period 6 and is in group two since it has two valence electrons. </em></u>
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Answer:

The  value  is  \lambda   = 1.329 *10^{-9} \  m

Explanation:

From the question we are told that

  The  orbital radius is  r =  0.846nm =  0.846 *10^{-9} \ m

Generally the de Broglie wavelength is mathematically represented as

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substituting values

     \lambda  =  \frac{ 2 * 3.142  *  0.846 *10^{-9}}{4}

    \lambda   = 1.329 *10^{-9} \  m

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