Kinetic energy of the emitted electron is given by
Te=ℏcλ−4.7>0T_e=\frac{\hbar c}{\lambda}-4.7>0Te=λℏc−4.7>0 λ=2000 :Te=1.58⋅10−19kg⋅m2s2−4.7⋅1.6⋅10−19kg⋅m2s2 =(1.58−7.52)⋅10−19J<0\lambda=2000 \colon T_e=1.58 \cdot 10^{-19} \frac{kg \cdot m^2}{s^2}-4.7 \cdot 1.6 \cdot 10^{-19} \frac{kg \cdot m^2}{s^2}\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=(1.58-7.52)\cdot 10^{-19} J<0λ=2000:Te=1.58⋅10−19s2kg⋅m2−4.7⋅1.6⋅10−19s2kg⋅m2=(1.58−7.52)⋅10−19J<0
Obviously, for λ=5000 :Te<0\lambda=5000 \colon T_e<0λ=5000:Te<0.
Hence, the energy of photons is too low for electrons to leave the surface.
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