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GuDViN [60]
3 years ago
9

For a grating, how many lines per millimeter would be required for the first order diffraction line for λ = 400 nm to be observe

d at a reflection angle of 5o when the angle of incidence is 45o?
Chemistry
1 answer:
MA_775_DIABLO [31]3 years ago
6 0

Answer:

For a grating, how many lines per millimeter would be required for the first order diffraction line for λ = 400 nm to be observed at a reflection angle of 5^{o} when the angle of incidence is  45^{o}?

<em>The number of lines per mm is 1985 lines </em>

Explanation:

A diffraction grating is used to separate a light source with multiple wavelengths into single wavelengths of different colors using the principle of diffraction.

Calculating for the spacing between the reflected surfaces

The spacing between the reflecting surfaces can be obtained using the relationship with the wavelength as shown below;

nA = d (sin i + sin r) ....................................1

where n is the order of diffraction = 1;

A is the wavelength = 400 nm

i is the angle of incidence = 45^{o}

r is the angle of refraction =  5^{o}

making d the subject formula from equation 1 we have;

d = \frac{nA}{sin i + sin r}...........................2

Now we substitute the given parameters into equation 2;

d = \frac{1(400 nm)}{(sin 45^{o}  ) + (sin 5^{o} )}

d = \frac{400 nm}{0.707 +0.087}

d = 503.7 nm

The spacing between reflecting surfaces is 503.7 nm

Calculating for the number of lines per millimeter

To calculate the number of spacing per millimeter we convert nanometer to millimeters, 1 mm =10^{6} and 1 line = 503.7 nm;

1 mm *\frac{1 line}{503.7 nm} *\frac{10^{6} }{mm}

= 1985 lines

Therefore the number of lines per millimeter would be 1985 lines

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<h3>What is a solution?</h3>

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6 0
2 years ago
A galvanic (voltaic) cell consists of an electrode composed of zinc in a 1.0 M zinc ion solution and another electrode composed
MariettaO [177]

Answer:

The E°cell for the galvanic cell is 1.56 V.

Explanation:

A galvanic cell is a device that uses redox reactions to convert chemical energy into electrical energy. The chemical reaction used is always spontaneous.

Oxide-reduction reactions, also called redox, involve the transfer or transfer of electrons between two or more chemical species. In these reactions two substances interact: the reducing agent and the oxidizing agent.

The gain of electrons is called reduction and the loss of electrons oxidation. That is to say, there is oxidation whenever an atom or group of atoms loses electrons (or increases its positive charges) and in the reduction an atom or group of atoms gains electrons, increasing its negative charges or decreasing the positive ones.

The species that supplies electrons is the reducing agent (that is, it is that species that oxidizes, yielding electrons and increasing its positive charge, or decreasing the negative one causing the reduction of the other species) and the one that gains them is the oxidizing agent ( that is, it is that species that is reduced, capturing electrons and increasing its negative charge, or decreasing its positive charge, causing oxidation of the other species).

The galvanic cell works as follows: In the anodic half-cell oxidations occur, while in the cathodic half-cell reductions occur. The anode electrode, conducts the electrons that are released in the oxidation reaction, to the metallic conductors. These electrical conductors conduct the electrons and carry them to the cathode electrode; the electrons thus enter the cathode half-cell and the reduction takes place in it.

To determine the oxidizing and reducing agent you must first know the reduction potentials. For this you consult the list of standard reduction potentials. In this list you can see that the semi-reactions that occur with their corresponding potentials are:

Ag⁺ + e⁻ ⇒ Ag E°= 0.80 V

Zn²⁺ + 2 e⁻ ⇒ Zn E° -0.76 V

The species that has the greatest potential for reduction will be the species that will be reduced, that is, it will be the oxidizing agent. In this case, it will be the experience corresponding to silver (Ag). Therefore, to obtain the redox reaction, the half-reaction corresponding to zinc (Zn) must be reversed to be an oxidation, keeping its E ° value constant. Then:

Reduction: Ag⁺ + e⁻ ⇒ Ag E°= 0.80 V

Oxidation: Zn ⇒ Zn²⁺ + 2 e⁻ E° -0.76 V

So: <em>E°cell=Ereduction - Eoxidation</em>

Or what is the same<em> E°cell=Ecathode - Eanode </em>because the reduction always occurs in the cathode and oxidation in the anode.

E°cell=0.80 V - (-0.76) V

<em>E°cell= 1.56 V</em>

Then <u><em>the E°cell for the galvanic cell is 1.56 V.</em></u>

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