Respuesta :
Explanation:
Given that,
Wavelength = 6.0 nm
de Broglie wavelength = 6.0 nm
(a). We need to calculate the energy of photon
Using formula of energy
[tex]E = \dfrac{hc}{\lambda}[/tex]
[tex]E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{6.0\times10^{-9}}[/tex]
[tex]E=3.315\times10^{-17}\ J[/tex]
(b). We need to calculate the kinetic energy of an electron
Using formula of kinetic energy
[tex]\lambda=\dfrac{h}{\sqrt{2mE}}[/tex]
[tex]E=\dfrac{h^2}{2m\lambda^2}[/tex]
Put the value into the formula
[tex]E=\dfrac{(6.63\times10^{-34})^2}{2\times9.1\times10^{-31}\times(6.0\times10^{-9})^2}[/tex]
[tex]E=6.709\times10^{-21}\ J[/tex]
(c). We need to calculate the energy of photon
Using formula of energy
[tex]E = \dfrac{hc}{\lambda}[/tex]
[tex]E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{6.0\times10^{-15}}[/tex]
[tex]E=3.315\times10^{-11}\ J[/tex]
(d). We need to calculate the kinetic energy of an electron
Using formula of kinetic energy
[tex]\lambda=\dfrac{h}{\sqrt{2mE}}[/tex]
[tex]E=\dfrac{h^2}{2m\lambda^2}[/tex]
Put the value into the formula
[tex]E=\dfrac{(6.63\times10^{-34})^2}{2\times9.1\times10^{-31}\times(6.0\times10^{-15})^2}[/tex]
[tex]E=6.709\times10^{-9}\ J[/tex]
Hence, This is the required solution.
