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anyanavicka [17]
2 years ago
14

How wide in m is a single slit that produces its first minimum for 624-nm light at an angle of 18. 0°?

Chemistry
1 answer:
Mariulka [41]2 years ago
6 0

A single slit with a width of 2019 * 109 m creates its initial minimal for 624 nm light at an angle of 18°.

How does diffraction work?

Waves spreading outward around obstructions are known as diffraction. Diffraction happens with sound, electromagnetic radiation like light, X-rays, and gamma rays, as well as with incredibly minuscule moving particles like atoms, neutrons, and electrons that exhibit wavelike qualities. Diffraction prevents the creation of sharp shadows as one of its effects. In order to spread out and illuminate regions at which a shadow is anticipated, light must be bent around corners, which is known as diffraction.

Calculation:

Provided for a single slit, m=1

λ = 624 *10⁻⁹

sinθ = sin 18⁰

Therefore,

asinθ=mλ

a = \frac{1 * 624 *10^{-9} }{sin 18}

⇒a = 2019 *10⁻⁹ m

Therefore the width of a single slit is 2019 *10⁻⁹ m.

Learn more about single slit here:

brainly.com/question/14283857

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A saturated solution of Al(OH)3 has a molar solubility of2.9x J0-9 Mat a certain temperature. What is the solubility product con
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<h3>What is solubility of product? </h3>

The solubility of product (Ksp) is defined as the concentration of products raised to their coefficient coefficients. This is illustrated below:

mA <=> nC + eD

Ksp = [C]^n × [D]^e

With the above information in mind, we can obtain the solubility of the product. This is illustrated below:

<h3>Dissociation equation </h3>

Al(OH)₃(aq) → Al³⁺(aq) + 3OH¯(aq)

From the balanced equation above,

1 mole of Al(OH)₃ contains 1 mole of Al³⁺ and 3 moles of OH¯

<h3>How to determine the concentration of Al³⁺ and OH¯</h3>

From the balanced equation above,

1 mole of Al(OH)₃ contains 1 mole of Al³⁺ and 3 moles of OH¯

Therefore,

2.9×10¯⁹ M Al(OH)₃ will also contain

  • 1 × 2.9×10¯⁹ = 2.9×10¯⁹ M Al³⁺
  • 3 × 2.9×10¯⁹ = 8.7×10¯⁹ M  OH¯

<h3>How to determine the solubility of product </h3>

Concentration of Al³⁺ = 2.9×10¯⁹ M

Concentration of OH¯ = 8.7×10¯⁹ M

Solubility product (Ksp) =?

Al(OH)₃(aq) → Al³⁺(aq) + 3OH¯(aq)

Ksp = [Al³⁺] × [OH¯]³

Ksp = 2.9×10¯⁹ × (8.7×10¯⁹)³

Ksp = 1.91×10¯³⁹

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