Answer:
λ = 65.6 pm
Explanation:
Given that
λo = 65 pm
The initial energy of the electron
[tex]E_o=\dfrac{hC}{\lambda_0 }[/tex]
Now by putting the values
[tex]E_o=\dfrac{hC}{\lambda_0 }[/tex]
[tex]E_o=\dfrac{6.67\times 10^{-34}\times 3\times 10^8}{65\times 10^{-12}}[/tex]
[tex]E_o=3.05\times 10^{-15}\ J[/tex]
[tex]E_o=\dfrac{3.05\times 10^{-15}}{1.6\times 10^{-19}\times 10^3}\ KeV[/tex]
Eo=19.06 KeV
Given that kinetic energy KE= 0.84 KeV
Therefore the final energy
E= Eo - KE
E = 19.06 - 0.84 KeV
E= 18.22 KeV
The wavelength λ can be find as
[tex]E=\dfrac{hC}{\lambda}[/tex]
[tex]\lambda=\dfrac{hC}{E}[/tex]
[tex]\lambda=\dfrac{6.67\times 10^{-34}\times 3\times 10^8}{19.06 \times 10^3\times 1.6\times 10^{-19}}[/tex]
λ = 6.56 x 10⁻¹¹ m
λ = 65.6 pm
