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tresset_1
28 days ago
15

The human eye can respond to as little as 10^−18 J of light energy. For a wavelength at the peak of visual sensitivity, 550 nm,

how many photons lead to an observable flash?
Physics
2 answers:
Yuliya22 [3.3K]28 days ago
7 0
The fundamental equation is derived from Mr. Planck: E=h \nu, where h is Planck’s constant and ν is the frequency. This relationship describes the energy per photon at a specific frequency. Although a wavelength is provided, it can easily be converted to frequency using the equation: c= lambda / nu, where c denotes the speed of light; λ (lambda) is the wavelength; and ν is the frequency. Once the energy of a photon with a wavelength of 550nm is determined, it will show how many photons are needed to gather 10^-18J. Remember to pay attention to the units.
ValentinkaMS [3.4K]28 days ago
6 0
The result is: 3 photons. To explain: The energy of a photon (E) can be formulated using: E = h*c/λ. Here, h represents Planck's constant (6.63x10⁻³⁴ J.s), λ signifies the wavelength (550 nm = 5.50x10⁻⁷m), and c stands for the speed of light (3.00x10⁸ m/s). Thus, E= (6.63x10⁻³⁴ * 3.00x10⁸)/(5.50x10⁻⁷) results in E = 3.62x10⁻¹⁹ J, which is the energy for one photon. Since the human eye can respond to 10⁻¹⁸ J, it follows that: 1 photon corresponds to 3.62x10⁻¹⁹ J, leading to a simple calculation resulting in approximately 2.76, which rounds to 3 photons.
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A person who weighs 625 N is riding a 98-N mountain bike. Suppose that the entire weight of the rider and bike is supported equa
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A = 4.76 x 10⁻⁴ m²

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2 months ago
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2 months ago
A 100-watt light bulb radiates energy at a rate of 100 J/s. (The watt, a unit of power or energy over time, is defined as 1 J/s.
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Answer:

2.64\times 10^{20} The number of photons emitted each second is

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Let 'n' stand for the quantity of photons released by the bulb.

Provided Information:

The bulb radiates energy at a rate of 100 J per second (E).

Wavelength of emitted light is (λ) = 525 nm = 525\times 10^{-9}\ m

The energy of a photon is calculated by:

Where,

E_0=\frac{hc}{\lambda}

Now, if we have 'n' photons, the total energy is equivalent to the energy of a single photon multiplied by the count of photons. Thus,

h\to Planck's\ constant=6.626\times 10^{-34}\ Js\\\\c\to Speed\ of \ light=3\times 10^{8}\ m/s

To express in terms of 'n', we find:

E=nE_0\\\\E=\frac{nhc}{\lambda}

Insert the provided values and solve for 'n'. The resulting calculation yields

n=\frac{E\lambda}{hc}

Consequently,

photons are discharged every second.n=\frac{100\times 525\times 10^{-9}}{6.626\times 10^{-34}\times 3\times 10^{8}}\\\\n=2.64\times 10^{20}

2.64\times 10^{20}

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