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3 years ago

Why is it better to use the metric system, rather than the English system, in scientific measurement?

Chemistry

2 answers:

nasty-shy [4]3 years ago

6 0

Answer:

Explanation:

Hello,

English system is indeed based on the metric system because they look for reference stuff to get the measurements, for instance, a foot equals 30.48 cm, but a common 30-cm rule measures every 1 cm, so the metric system also provides more accurate measurements. An other example lies on the measurement of mass, a pound which equals 453.59 g is appropriated to measure middle amounts of masses, but for example, in qualitative analysis we tend to measure amounts below about 0.10 g (0.022 lb) so it is easier for us in science to use the metric system in addition to the straightforwardness of the reporting data in 10-based notation (scientific notation).

Best regards.

frez [133]3 years ago

3 0

The metric system is based on powers of 10 so it is much easier to convert units, often just by moving the decimal point.

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an electron in the 3rd shell of an Aluminium atom moves to the first shell in a bombardment process. Calculate the frequency of

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Following the quantic theory, the energy of a photon equals the radiation frequency multiplied by the universal constant. ν = 2.923x10¹⁵ Hz. E = 3.09x10¹⁵Hz.

<h3>What is quantum mechanic?</h3>

It is the branch of physics that studies objects and forces at a very low scale, at atoms, subatoms, and particles levels.

Quantum mechanics states that the elemental particles that constitute matter -electrons, neutrons, protons- have the properties of a wave and a particle.

It emerges from the quantic theory exposed by Max Planck (1922), in which he affirmed that light propagates in energy packages or photons.

He discovered the Universal Planck constant, h, used to calculate the energy of a photon.

He stated that the energy of a photon (E) equals the radiation frequency (ν) multiplied by the universal constant (h).

E = νh

In the exposed example, we need to calculate the energy required to change from the 3rd shell to the first shell.

To do it, we should know that the energy in a level (Eₙ) equals the energy associated to an electron in the most inferior energy level (E₁) divided by the square of the shell number (n²).

Eₙ = E₁ / n²

E₁ is a constant. We can express it in Joules or electroVolts

E₁ = -2.18x10⁻¹⁸ J
E₁ = -13.6 eV

So, let us calculate the energy at level 1 and 3

Eₙ = E₁ / n²

E₁ = -2.18x10⁻¹⁸ J / 1² = -2.18x10⁻¹⁸ J

E₁ = -13.6 eV / 1² = -13.6 eV

E₃ = -2.18x10⁻¹⁸ J / 3² = -2.18x10⁻¹⁸ J / 9 = - 2.42x10⁻¹⁹ J

E₃ = -13.6 eV / 3² = -13.6 eV / 9 = - 1.51 eV

The change of energy can be calculated in two ways,

Option 1

ΔE = E₁ - E₃ = 2.18x10⁻¹⁸ - 2.42x10⁻¹⁹ = 1.93x10⁻¹⁸J

ΔE = E₁ - E₃ = 13.6 - 1.51 = 12.09 eV

Option 2

ΔE = -2.18x10⁻¹⁸ J (1/nf² - 1/ni²)

ΔE =-13.6 eV (1/nf² - 1/ni²)

Where nf is the final level and ni is the initial level. When the electron passes from its initial level to its final level it is called electronic transition.

ni = 3
nf = 1

ΔE = -2.18x10⁻¹⁸ J (1/nf² - 1/ni²)

ΔE = -2.18x10⁻¹⁸ J (1/1² - 1/3²)

ΔE = -2.18x10⁻¹⁸ J (1 - 0.111)

ΔE = -2.18x10⁻¹⁸ J (0.888)

ΔE = - 1.937x10⁻¹⁸ J

ΔE = -13.6 eV (1/nf² - 1/ni²)

ΔE = -13.6 eV (1/1² - 1/3²)

ΔE = -13.6 eV (1 - 0.111)

ΔE = -13.6 eV (0.888)

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This is the energy required for the electron to go from n= 3 to n = 1. The negative sign (-) means energy (as light or photons) released or emitted.

If we want to express the result in Hz, we just need to make a conversion.

1Hz ⇔ 6.626x10⁻³⁴J ⇔ 4.136x10¹⁵ eV.

The energy required for the electron to go from n= 3 to n = 1 is 3.09x10¹⁵ Hz.

Now, we need to calculate the frequency, ν. This is, how many times the wave oscillates back and foward per second.

To do it, we will use the universal Planck constant, h, and the absolute value of the energy, E.

ν = E/h = 1.937x10⁻¹⁸ J / 6.626x10⁻³⁴ Js = 2.923x10¹⁵ 1/s = 2.923x10¹⁵ Hz.

Answer:

Frequency, ν = E/h = 2.923x10¹⁵ Hz.
Energy, E = 3.09x10¹⁵ Hz.

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