Question #69998

Calculate the wavelength of the light required to eject a photoelectron from caesium metal with a kinetic energy of 2.0 x 10 ^-19 J (v° for caesium is 4.55 x 10^14 Hz.

Expert's answer

Answer on Question # 69998, Chemistry / Inorganic Chemistry

Calculate the wavelength of the light required to eject a photoelectron from caesium metal with a kinetic energy of 2.0×10−19 J2.0 \times 10^{-19} \, \text{J} (v for caesium is 4.55×1014 Hz4.55 \times 10^{14} \, \text{Hz})?

Solution:

1. Calculate the general energy:


Kmax⁡=h×f−Φ0K_{\max} = h \times f - \Phi_0Kmax⁡=6,63×10−34×4.55×1014−2.0×10−19=1.0×10−19 (J)K_{\max} = 6,63 \times 10^{-34} \times 4.55 \times 10^{14} - 2.0 \times 10^{-19} = 1.0 \times 10^{-19} \, (J)


2. Calculate the wavelength:


Kmax⁡=h×cλ⇒λ=h×cKmax⁡K_{\max} = \frac{h \times c}{\lambda} \Rightarrow \lambda = \frac{h \times c}{K_{\max}}λ=6,63×10−34×3×1081.0×10−19=198.9×10−8 (m)\lambda = \frac{6,63 \times 10^{-34} \times 3 \times 10^8}{1.0 \times 10^{-19}} = 198.9 \times 10^{-8} \, (m)


Answer: 198.9×10−8 (m)198.9 \times 10^{-8} \, (m)

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