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disa [49]
1 year ago
7

In his explanation of the threshold frequency in the photoelectric effect, Einstein reasoned that the absorbed photon must have

a minimum energy to dislodge an electron from the metal surface. This energy is called the work function (Ф) of the metal. What is the longest wavelength of radiation (in nm) that could cause the photoelectric effect in each of these metals:
(a) calcium, Ф=4.60 × 10⁻¹⁹J
Chemistry
1 answer:
astraxan [27]1 year ago
7 0

The wavelength of radiation that could cause the photoelectric effect in calcium is &\lambda=4.32×10^{-7} m

<h3>What is the photoelectric effect?</h3>

The photoelectric effect is a phenomenon that results in electrically charged particles being discharged from or within a substance when it absorbs electromagnetic radiation. When light strikes a metal plate, the action is frequently described as the ejection of electrons from the plate.

For calcium, the wavelength will be

$$\begin{aligned}&\mathrm{E}=4.6^{*} 10^{-19} \mathrm{~J} \\&\mathrm{~h}=6.626^{*} 10^{-34} \mathrm{Js} \\&\mathrm{c}=3^{*} 10^{8} \mathrm{~m} / \mathrm{s} \\&\mathrm{E}=\frac{h c}{\lambda} \\&\lambda=\frac{h c}{E} \\&\lambda=\frac{6.626 * 10^{-34} \mathrm{Js} * 3 * 10^{8} \mathrm{~m} / \mathrm{s}}{4.6 * 10^{-19} \mathrm{~J}} \\&\lambda=4.32^{*} 10^{-7} \mathrm{~m}\end{aligned}$$

The wavelength of radiation that could cause the photoelectric effect in calcium is &\lambda=4.32 ×10^{-7} \mathrm{~m}.

To know more about the photoelectric effect, visit: brainly.com/question/1279581

#SPJ4

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<h3>Answer:</h3>

0.024 kg CaO

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Aqueous Solutions</u>

  • Molarity = moles of solute / liters of solution

<u>Atomic Structure</u>

  • Reading a Periodic Tables
  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

0.41 mol CaO

2.5 M Solution

<u>Step 2: Identify Conversions</u>

1000 g = 1 kg

Molar Mass of Ca - 40.08 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of CaO - 40.08 + 16.00 = 56.08 g/mol

<u>Step 3: Convert</u>

  1. Set up:                                \displaystyle 0.43 \ mol \ CaO(\frac{56.08 \ g \ Cao}{1 \ mol \ CaO})(\frac{1 \ kg \ CaO}{1000 \ g \ CaO})
  2. Multiply:                              \displaystyle 0.024114 \ kg \ CaO

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs as our lowest.</em>

0.024114 kg CaO ≈ 0.024 kg CaO

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