The number of photons that were absorbed by the water is about:
1.00 × 10²⁹ photons
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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:
[tex]\large {\boxed {E = h \times f}}[/tex]
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.
[tex]\large {\boxed {E = \frac{1}{2}mv^2 + \Phi}}[/tex]
[tex]\large {\boxed {E = qV + \Phi}}[/tex]
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 = λ = 14.1 cm
volume of water = V = 0.485 L
change in temperature = Δt = 69.9 °C
density of water = ρ = 1.00 g/mL = 1.00 kg/L
specific heat capacity of water = c = 4.184 J/g°C = 4184 J/kg°C
speed of light = c = 3 × 10⁸ m/s
Asked:
number of photons = N = ?
Solution:
[tex]\texttt{ Heat absorbed by water } = \texttt{ Total Energy of Photons }[/tex]
[tex]m c \Delta t = N h f[/tex]
[tex]\rho V c \Delta t = N h \frac{c}{\lambda}[/tex]
[tex]N = ( \rho V c \Delta t ) \div ( h \frac{c}{\lambda} )[/tex]
[tex]N = ( 1.00 \times 0.485 \times 4184 \times 69.9 ) \div ( 6.63 \times 10^{-34} \times \frac{3 \times 10^8}{ 0.141 } )[/tex]
[tex]N \approx 1.00 \times 10^{29} \texttt{ photons}[/tex]
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Learn more
- Photoelectric Effect : https://brainly.com/question/1408276
- Statements about the Photoelectric Effect : https://brainly.com/question/9260704
- Rutherford model and Photoelecric Effect : https://brainly.com/question/1458544
- Photoelectric Threshold Wavelength : https://brainly.com/question/10015690
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Answer details
Grade: High School
Subject: Physics
Chapter: Quantum Physics
