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yanshen's Question
Physics E&M
Posted 2 months ago

・波長 λ\lambda305 nm305 \mathrm{~nm} 電磁波のフォトンは,
1.77 eVの運動エネルギーをもった電子をある金属から放出させる。 このとき,金属から電子を放出させることのできる電磁波の 最大波長を計算せよ。
ただし,計算には下の数値・関係を用いよ。
プランク定数 h=6.626×1034Jsh=6.626 \times 10^{-34} \mathrm{Js}
光の速度 c=2.998×108 ms1c=2.998 \times 10^{8} \mathrm{~ms}^{-1}
振動数と波長の関係 v=c/λv=c / \lambda
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Answer from Sia
Posted 2 months ago
Answer
The maximum wavelength of electromagnetic waves that can cause the emission of electrons from the metal is approximately 700 nm.
Solution
a
Given Data: The wavelength λ\lambda of the electromagnetic wave is 305nm305 \, \text{nm}, and the kinetic energy of the emitted electron is 1.77eV1.77 \, \text{eV}. The Planck constant hh is 6.626×1034Js6.626 \times 10^{-34} \, \text{Js}, and the speed of light cc is 2.998×108m/s2.998 \times 10^8 \, \text{m/s}
b
Energy of Photon: The energy of a photon EE can be calculated using the formula E=hcλE = \frac{hc}{\lambda}. Substituting the given values, we get: E=6.626×1034Js×2.998×108m/s305×109m6.51×1019J E = \frac{6.626 \times 10^{-34} \, \text{Js} \times 2.998 \times 10^8 \, \text{m/s}}{305 \times 10^{-9} \, \text{m}} \approx 6.51 \times 10^{-19} \, \text{J} Converting this energy to electron volts (1 eV = 1.602×1019J1.602 \times 10^{-19} \, \text{J}): E6.51×1019J1.602×1019J/eV4.06eV E \approx \frac{6.51 \times 10^{-19} \, \text{J}}{1.602 \times 10^{-19} \, \text{J/eV}} \approx 4.06 \, \text{eV}
c
Work Function: The work function ϕ\phi of the metal can be found using the photoelectric equation E=ϕ+K.E.E = \phi + K.E., where K.E.K.E. is the kinetic energy of the emitted electron. Rearranging for ϕ\phi: ϕ=EK.E.=4.06eV1.77eV=2.29eV \phi = E - K.E. = 4.06 \, \text{eV} - 1.77 \, \text{eV} = 2.29 \, \text{eV}
d
Maximum Wavelength: The maximum wavelength λmax\lambda_{\text{max}} that can cause electron emission is found using ϕ=hcλmax\phi = \frac{hc}{\lambda_{\text{max}}}. Rearranging for λmax\lambda_{\text{max}}: λmax=hcϕ=6.626×1034Js×2.998×108m/s2.29×1.602×1019J5.43×107m=543nm \lambda_{\text{max}} = \frac{hc}{\phi} = \frac{6.626 \times 10^{-34} \, \text{Js} \times 2.998 \times 10^8 \, \text{m/s}}{2.29 \times 1.602 \times 10^{-19} \, \text{J}} \approx 5.43 \times 10^{-7} \, \text{m} = 543 \, \text{nm}
Key Concept
The photoelectric effect and the relationship between photon energy, work function, and kinetic energy of emitted electrons.
Explanation
The maximum wavelength of electromagnetic waves that can cause the emission of electrons from a metal is determined by the work function of the metal and the energy of the incident photons.

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