Chemistry, 30.01.2020 06:49 dondre54

The human eye is able to detect as little as 2.35 × 10–18 j of green light of wavelength 510 nm. calculate the minimum number of photons of green light that can be detected by the human eye. physical constants can be found here.

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Answer from: janaemartinez42

The minimum number of photons that can be detected by the human eye is 6.03 × 10 ¹⁶ photons.

Explanation:

The energy of one photon of light is related to the wavelength by the equation:

E = h×c/λ

Where, E is the energy of one photon, h is the Planck's constant, c is the speed of light, and λ is the wavelength of the light.

You are given with λ = 510 nm (nanometers), which you must convert to m (meters), to use SI units ⇒ λ = 510 × 10⁻⁹ m.

The physical constansts needed are:

Planck's constant, h = 6.63 × 10⁻³⁴ J.sSpeed of light, in vacuum, c = 3.0 × 10⁻⁸ m/s

Now you can substitute in the formula can compute for the value of E:

E = 6.63×10⁻³⁴ J.s × 3.0 × 10⁻⁸ m/s / (510×10⁻⁹ m) = 0.039 × 10⁻³³ J

Since, that is the energy of one photon of green light, to calculate the number of photons that can be detected by the human eye, you need to divide the amounf of energy the human eye is able to detect, 2.35 × 10⁻¹⁸ J , by the energy of a photon:

number of photons = 2.35×10 ⁻¹⁸J / 0.039 × 10⁻³³ J/ photonnumber of photons = 60.3 × 10¹⁵ photons = 6.03 × 10¹⁶photons

Answer from: princess6039

The minimum number of photons of green light, with 2.35x10⁻¹⁸ J of energy and 510 nm of wavelength, that can be detected by the human eye is 6.04 photons.

The energy of a single photon is given by:

$E_{1} = h\frac{c}{\lambda}$

Where:

h: is the Planck constant = 6.62x10⁻³⁴ J*s

λ: is the wavelength of the green light = 510 nm = 510x10⁻⁹ m

c: is the speed of light = 3.00x10⁸ m/s

So, the energy of one photon is:

$E_{1} = h\frac{c}{\lambda} = 6.62 \cdot 10^{-34} Js\frac{3.00 \cdot 10^{8} m/s}{510 \cdot 10^{-9} m} = 3.89 \cdot 10^{-19} J$

Now, since the energy of the green light detected by the human eye is 2.35x10⁻¹⁸ J, the minimum number of photons (n) to produce that amount of energy is:

$E_{n} = nE_{1}$

$n = \frac{E_{n}}{E_{1}} = \frac{2.35 \cdot 10^{-18} J}{3.89 \cdot 10^{-19} J} = 6.04$

Therefore, the minimum number of photons of green light is 6.04 photons.

You can find more about energy here:

linklink

I hope it helps you!

The human eye is able to detect as little as 2.35 × 10–18 j of green light of wavelength 510 nm. cal

Answer from: Quest

here we have to get the change of moles on decrease the volume of the reaction in gas phase.

the per the given options it will decrease by (d) 11-50%

we know, on decreasing the volume of the gas phase reaction the reaction moves to the product side.

the reaction: so₂cl₂ (g) = so₂ (g) + cl₂ (g) total

the moles at equilibrium: 1 - α α α = 1+α

concentrations: 1 - α α α = 1 + α

(concentration = moles/volume, as volume is 1 l)

(where α is the degree of dissociation)

now the dissociation constant in terms of concentration is :

$k_{c}$ = $\frac{[so2cl2] x [cl2]}{[so2cl2]}$

or, $\frac{\alpha ^{2} }{(1-\alpha)v }$

or, $k_{c}$ = $\frac{\alpha ^{2} }{(1-\alpha) }$ (at v = 1l)

the reaction: so₂cl₂ (g) = so₂ (g) + cl₂ (g) total

the moles at equilibrium: 1 - α α α = 1+α

concentrations: (1 - α)/0.5 α/0.5 α/0.5 = (1 + α)/0.5

thus, the dissociation constant in terms of concentration is :

$\frac{[so2cl2] x [cl2]}{[so2cl2]}$

or, $\frac{\alpha ^{2} }{(1-\alpha)0.5 }$

$\frac{\alpha ^{2} }{(1-\alpha) }$ = $\frac{\alpha ^{2} }{(1-\alpha) 0.5}$

so, the change of 1 : 1/0.5

or, 1: 2

thus it will decrease by 50%.

or, as per the given options it will decrease by 11-50%

Answer from: Quest

since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.

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