・波長 $\lambda$ が $305 \mathrm{~nm}$ 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 $h=6.626 imes 10^{-34} \mathrm{Js}$
光の速度 $c=2.998 imes 10^{8} \mathrm{~ms}^{-1}$
振動数と波長の関係 $v=c / \lambda$

Question

・波長 $\lambda$ が $305 \mathrm{~nm}$ 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 $h=6.626 	imes 10^{-34} \mathrm{Js}$
光の速度 $c=2.998 	imes 10^{8} \mathrm{~ms}^{-1}$
振動数と波長の関係 $v=c / \lambda$

Accepted Answer

∻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 $\lambda$ of the electromagnetic wave is $305 \, \text{nm}$, and the kinetic energy of the emitted electron is $1.77 \, \text{eV}$. The Planck constant $h$ is $6.626 \times 10^{-34} \, \text{Js}$, and the speed of light $c$ is $2.998 \times 10^8 \, \text{m/s}$. ‖ ‖ b ⋮ Energy of Photon: The energy of a photon $E$ can be calculated using the formula $E = \frac{hc}{\lambda}$. Substituting the given values, we get: $$ E = \frac{6.626 \times 10^{-34} \, \text{Js} \times 2.998 \times 10^8 \, \text{m/s}}{305 \times 10^{-9} \, \text{m}} \approx 6.51 \times 10^{-19} \, \text{J} $$ Converting this energy to electron volts (1 eV = $1.602 \times 10^{-19} \, \text{J}$): $$ E \approx \frac{6.51 \times 10^{-19} \, \text{J}}{1.602 \times 10^{-19} \, \text{J/eV}} \approx 4.06 \, \text{eV} $$ ‖ ‖ c ⋮ Work Function: The work function $\phi$ of the metal can be found using the photoelectric equation $E = \phi + K.E.$, where $K.E.$ is the kinetic energy of the emitted electron. Rearranging for $\phi$: $$ \phi = E - K.E. = 4.06 \, \text{eV} - 1.77 \, \text{eV} = 2.29 \, \text{eV} $$ ‖ ‖ d ⋮ Maximum Wavelength: The maximum wavelength $\lambda_{\text{max}}$ that can cause electron emission is found using $\phi = \frac{hc}{\lambda_{\text{max}}}$. Rearranging for $\lambda_{\text{max}}$: $$ \lambda_{\text{max}} = \frac{hc}{\phi} = \frac{6.626 \times 10^{-34} \, \text{Js} \times 2.998 \times 10^8 \, \text{m/s}}{2.29 \times 1.602 \times 10^{-19} \, \text{J}} \approx 5.43 \times 10^{-7} \, \text{m} = 543 \, \text{nm} $$ ‖ ∻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. ⚹◊What is the formula relating energy to frequency for a photon in the electromagnetic spectrum?⍭ Generate me a similar question◊

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