Question

How many photons per second are emitted by a monochromatic lightbulb (650 nm) that emits 45 W of power? Express your answer using two significant figures. If you stand 17 m from this bulb, how many photons enter each of your eyes per second? Assume your pupil is 5.0 mm in diameter and that the bulb radiates uniformly in all directions. Express your answer using two significant figures.

Answer 1

Answer:

The number of photons per second are $7.95\times10^{11}\ photons/s$.

Explanation:

Given that,

Wavelength = 650 nm

Power = 45 W

Distance R= 17 m

Diameter = 5.0 mm

We need to calculate the number of photon per second emitted by light bulb

Using formula of energy

$E=\dfrac{hc}{\lambda}$

The power is

$P=\dfrac{nE}{t}$

$\dfrac{n}{t}=\dfrac{P}{E}$

Put the value of E

$\dfrac{n}{t}=\dfrc{P \lambda}{hc}$

Put the value into the formula

$\dfrac{n}{t}=\dfrac{45\times650\times10^{-9}}{6.6\times10^{-34}\times3\times10^{8}}$

$\dfrac{n}{t}=1.47\times10^{20}\ photons/s$

We need to calculate the surface area

Using formula of area

$A=4\piR^2$

$A=4\pi\times17^2$

We need to calculate the number of photons entering into eye

$N=n\dfrac{A_{eye}}{A_{surface}}$

$N=1.47\times10^{20}\times\dfrac{\pi(2.5\times10^{-3})^2}{4\pi\times17^2}$

$N=7.95\times10^{11}\ photons/s$

Hence, The number of photons per second are $7.95\times10^{11}\ photons/s$.

Answer 2

A. 1.5 × 10²⁰ photons per second are emitted by a monochromatic lightbulb

B. 8.0 × 10¹¹ photons per second enter each of the eyes

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Further explanation

The term of package of electromagnetic wave radiation energy was first introduced by Max Planck. He termed it with photons with the magnitude is :

$\large {\boxed {E = h \times f}}$

E = Energi of A Photon ( Joule )

h = Planck's Constant ( 6.63 × 10⁻³⁴ Js )

f = Frequency of Eletromagnetic Wave ( Hz )

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The photoelectric effect is an effect in which electrons are released from the metal surface when illuminated by electromagnetic waves with large enough of radiation energy.

$\large {\boxed {E = \frac{1}{2}mv^2 + \Phi}}$

$\large {\boxed {E = qV + \Phi}}$

E = Energi of A Photon ( Joule )

m = Mass of an Electron ( kg )

v = Electron Release Speed ( m/s )

Ф = Work Function of Metal ( Joule )

q = Charge of an Electron ( Coulomb )

V = Stopping Potential ( Volt )

Let us now tackle the problem !

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Given:

wavelength of the light = λ = 650 nm = 6.5 × 10⁻⁷ m

power of lightbulb = P = 45 W

distance between the eyes and the bulb = R = 17 m

diameter of pupil = d = 5.0 mm = 5.0 × 10⁻³ m

Asked:

A. total number of photons per second = N/t = ?

B. number of photon per second entering the eyes = n/t = ?

Solution:

Part A:

$P = E \div t$

$P = N h f \div t$

$P = \frac{N}{t} hf$

$\frac{N}{t} = P \div (hf}$

$\frac{N}{t} = P \div (h\frac{c}{\lambda})$

$\frac{N}{t} = 45 \div (6.63 \times 10^{-34} \times \frac{3 \times 10^8}{6.5 \times 10^{-7}})$

$\boxed{\frac{N}{t} \approx 1.5 \times 10^{20} \texttt{ photons/second}}$

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Part B:

$\frac{n}{t} = \frac{A_{eye}}{A_{total}} \times \frac{N}{t}$

$\frac{n}{t} = \frac{\frac{1}{4} \pi d^2}{4 \pi R^2} \times \frac{N}{t}$

$\frac{n}{t} = \frac{d^2}{16 R^2} \times \frac{N}{t}$

$\frac{n}{t} = \frac{(5.0 \times 10^{-3})^2}{16 \times 17^2} \times 1.5 \times 10^{20}$

$\boxed{\frac{n}{t} = 8.0 \times 10^{11} \texttt{ photons/second} }$

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Learn more

• Photoelectric Effect : brainly.com/question/1408276
• Statements about the Photoelectric Effect : brainly.com/question/9260704
• Rutherford model and Photoelecric Effect : brainly.com/question/1458544

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Answer details

Grade: College

Subject: Physics

Chapter: Quantum Physics