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Sagot :
Answer: [tex]140\ nm[/tex]
Explanation:
Given
Initial kinetic energy is [tex]K_i=13.2\ eV[/tex]
Final kinetic energy is [tex]K_f=4.5\ eV[/tex]
Change in kinetic energy is [tex]\Delta K=K_i-K_f[/tex]
[tex]\Rightarrow \Delta K=13.2-4.5\\\Rightarrow \Delta K=8.7\ eV[/tex]
Energy is given by
[tex]E=\dfrac{hc}{\lambda}[/tex]
Equate it to change in kinetic energy
[tex]\Rightarrow 8.7=\dfrac{hc}{\lambda}\\\\\Rightarrow \lambda =\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{8.7\times 1.6\times 10^{-19}}\\\\\Rightarrow \lambda =1.4\times 10^{-7}\ m\\\Rightarrow \lambda=140\ nm[/tex]
Therefore, the wavelength associated with it is [tex]140\ nm[/tex].
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