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Minchanka [31]
3 years ago
15

What is the minimum frequency of light required to observe the photoelectric effect on pd

Chemistry
1 answer:
Tems11 [23]3 years ago
3 0

Answer:

1.26 × 10¹⁵ s⁻¹

Explanation:

Work function is the minimum energy required to remove an electron from the surface of metal

energy of the electron = hf - Φ

Φ = work function = hf₀ where f₀  = threshold frequency

f₀ = Φ / h  where h ( Planck constant = 6.626 × 10⁻³⁴ Js)

Φ = 5.22eV = 5.22 × 1 eV  where 1 eV = 1.60217662 × 10⁻¹⁹ J

Φ = 5.22 × 1.60217662 × 10⁻19 J = 8.363362 × 10⁻¹⁹ J

f₀ =  (8.363362 ×10⁻¹⁹  J) / (6.626× 10⁻³⁴ Js) = 1.26 × 10¹⁵ s⁻¹

The frequency must be greater than the 1.26 × 10¹⁵ s⁻¹ to observe the emission

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Write a chemical equation representing the first ionization energy for lithium. use e− as the symbol for an electron.
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The chemical equation representing the first ionization energy for lithium is given by;

Li → Li + e-

<h2>Further Explanation; </h2><h3>Ionization energy</h3>
  • Ionization energy is the energy required to remove outermost electrons from the outermost energy level. Energy is required to remove an electron from an atom.
  • The closer an electron is to the nucleus the more energy is required, since the electron is more tightly bound to the atom thus making it more difficult to remove, hence higher ionization energy.
  • Ionization energy increases across the periods and decreases down the group from top to bottom.  
  • Additionally, the ionization energy increases with subsequent removal of a second or a third electron.
<h3>First ionization energy  </h3>
  • This is the energy required to remove the first electron from the outermost energy level of an atom.
  • Energy needed to remove the second electron to form a divalent cation is called the second ionization energy.
<h3>Trends in ionization energy  </h3><h3>1. Down the group(top to bottom)</h3>
  • Ionization energy decreases down the groups in the periodic table from top to bottom.
  • It is because as you move down the group the number of energy levels increases making the outermost electrons get further from the nucleus reducing the strength of attraction to the nucleus.
  • This means less energy will be required compared to an atoms of elements at the top of the groups.
<h3>2. Across the period  (left to right)</h3>
  • Ionization energy increases across the period from left to right.
  • This can be explained by an increase in nuclear energy as extra protons are added to the nucleus across the period increasing the strength of attraction of electrons to the nucleus.
  • Consequently, more energy is needed to remove electrons from the nucleus.

Keywords: Ionization energy, periodic table, energy levels, electrons

<h3>Learn more about</h3>
  • Ionization energy: brainly.com/question/1971327
  • Trend in ionization energy: brainly.com/question/1971327
  • First ionization energy: brainly.com/question/1971327

Level: High school  

Subject: Chemistry  

Topic: Periodic table and chemical families  

Sub-topic: Ionization energy

4 0
3 years ago
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What type of reaction occurs in a hand warmer?<br> Exothermic<br> Endothermic
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The reaction is exothermic.

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8 0
2 years ago
Which reaction will occur? 2NaBr+I2 —&gt; 2NaI + Br2
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If a particular ore contains 58.6 % calcium phosphate, what minimum mass of the ore must be processed to obtain 1.00 kg of phosp
rjkz [21]
Answer is: mass of the ore is 8.54kg.<span>

</span>ω(Ca₃(PO₄)₂ - calcium phosphate) = 58.6% ÷ 100% = 0.586.
m(P) = 1.00 kg · 1000 g/kg.
m(P) = 1000 g.
In one molecule of calcium phosphate there are two phosphorus atoms:
M(Ca₃(PO₄)₂) = 310.18 g/mol.
M(P) = 30.97 g/mol.
For one kilogram of phosphorus, we need:
M(Ca₃(PO₄)₂) : 2M(P) = m(Ca₃(PO₄)₂) : m(P).
310.18 g/mol : 61.94 g/mol = m(Ca₃(PO₄)₂) : 1000 g.
m(Ca₃(PO₄)₂) = 5007.75 g ÷ 1000 g/kg = 5.007 kg.
Mass of ore find from proportion:
m(Ca₃(PO₄)₂) : m(ore) = 56% : 100%.
m(ore) = 100% · 5.007 kg ÷ 58.6%.
m(ore) = 8.54kg.

5 0
3 years ago
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