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inna [77]
3 years ago
7

The human eye is a complex sensing device for visible light. The optic nerve needs a minimum of 2.0 × 10−17 J of energy to trigg

er a series of impulses that eventually reach the brain. How many photons of blue light (475 nm) are needed?

Chemistry
2 answers:
Karo-lina-s [1.5K]3 years ago
7 0
Number of photons can be calculated by dividing the needed energy by the energy per photon.

The minimum energy needed is given as 2 x 10^-17 joules
Energy per photon = hc / lambda where h is planck's constant, c is the speed of light and lambda is the wavelength
Energy per photon = (<span>6.626 x 10^-34 x 3 x 10^8) / (475 x 10^-9)
                              = 4.18 x 10^-19 J

number of photons = (2 x 10^-17) / (4.18 x 10^-19)
                               = 47.79 photons which is approximately 48 photons</span>
Sonja [21]3 years ago
5 0

The number of photons needed to get the energy to trigger a series of impulses that eventually reaches the brain is: 48.7 photons

<h3><em>Further explanation</em></h3>

The photoelectric effect is an electron coming out of a metal because of electromagnetic radiation

One type of electromagnetic radiation is light

Electrons can come out of metal because they absorb electromagnetic energy radiated on metals. There is also kinetic energy released from metal, which is according to the equation:

E = hf - hfo

fo = the threshold frequency of electromagnetic waves

Radiation energy is absorbed by photons

The energy in one photon can be formulated as

\large{\boxed{\bold{E\:=\:h\:.\:f}}}

Where

h = Planck's constant (6,626.10⁻³⁴ Js)

f = Frequency of electromagnetic waves

f = c / λ

c = speed of light

= 3.10⁸

λ = wavelength

Known wavelength 475 nm

475 nm = 475.10⁻⁹ m

We put it in the formula to find this light energy E = h. f

E = h. c /λ

E =: 6,626.10⁻³⁴. 3.10⁸ / 475 .10⁻⁹ m

E = 4.1.10⁻¹⁹

Because this is 1 photon energy, while the minimum energy required is 2.0 × 10⁻¹⁷ J, the number of photons needed is:

number of photons = 2.0 × 10−17 J: 4.1.10⁻¹⁹

number of photons = 48.7

<h3><em>Learn more</em></h3>

Electromagnetic radiation with wavelength of 745 nm

brainly.com/question/7590814

The energy of a photon

brainly.com/question/7353559

The equation E = hf

brainly.com/question/4177755

the approximate energy of a photon

brainly.com/question/7991589

The equation for photon energy

brainly.com/question/2741868

the wavelength of a photon whose energy is twice

brainly.com/question/6576580

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a 1.25g sample of ore containing iron pyrite (FeS2) was pulverized and ignited in air, converting the FeS2 to Fe2O3 and SO2(g).
svp [43]

Answer:

28.9%

Explanation:

Let's consider the following balanced equation.

2 FeS₂ + 11/2 O₂ ⇒ Fe₂O₃ + 4 SO₂

We can establish the following relations:

  • The molar mass of Fe₂O₃ is 159.6 g/mol
  • 1 mole of Fe₂O₃ is produced per 2 moles of FeS₂
  • 1 mole of Fe is in 1 mole of FeS₂
  • The molar mass of Fe is 55.84 g/mol

The amount of Fe in the sample that produced 0.516 g of Fe₂O₃ is:

0.516gFe_{2}O_{3}.\frac{1molFe_{2}O_{3}}{159.6gFe_{2}O_{3}} .\frac{2molFeS_{2}}{1molFe_{2}O_{3}} .\frac{1molFe}{1molFeS_{2}} .\frac{55.84gFe}{1molFe} =0.361gFe

The percent of Fe in 1.25 g of the ore is:

\frac{0.361g}{1.25g} .100\%=28.9\%

4 0
3 years ago
How many atoms are in 12 moles of silver?
lord [1]

Answer:

NA=72.264x1023

Explanation:

I got you

7 0
3 years ago
How many grams are there in 2.75E23 molecules of C2H6 (MM = 30 grams)?
skelet666 [1.2K]

Answer:13.70 i think

Explanation:

3 0
2 years ago
It the mass of a material is 46 grams and the volume of the material is 8 cm ^3 What would the density of the material to be
Nat2105 [25]
5.75 Grams per cm^3

You do mass divided by volume
3 0
3 years ago
Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and is allowed to reach equilibriumdescribed by the equation N2O4(g)↔
Vilka [71]

Answer : The correct option is, (C) 1.1

Solution :  Given,

Initial moles of N_2O_4 = 1.0 mole

Initial volume of solution = 1.0 L

First we have to calculate the concentration N_2O_4.

\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}

\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M

The given equilibrium reaction is,

                           N_2O_4(g)\rightleftharpoons 2NO_2(g)

Initially                      c                 0

At equilibrium   (c-c\alpha)           2c\alpha

The expression of K_c will be,

K_c=\frac{[NO_2]^2}{[N_2O_4]}

K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}

where,

\alpha = degree of dissociation = 40 % = 0.4

Now put all the given values in the above expression, we get:

K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}

K_c=\frac{(2\times 1\times 0.4)^2}{(1-1\times 0.4)}

K_c=1.066\aprrox 1.1

Therefore, the value of equilibrium constant for this reaction is, 1.1

4 0
3 years ago
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