Respuesta :
Answer:
The recoil speed is [tex]2.207\times 10^{4} m/s[/tex]
Solution:
Wavelength of a blue-green photon, [tex]\lambda_{BG} = 488 nm = 488\times 10^{- 9} m[/tex]
Now, the energy associated with the blue-green photon:
[tex]E_{BG} = \frac{hc}{\lambda_{BG}}[/tex]
where
h = Planck's constant
C = speed of light ion vacuum
[tex]E_{BG} = \frac{6.626\times 10^{- 34}\times 3\times 10^{8}}{488\times 10^{- 9}}[/tex]
[tex]E_{BG} = 4.07\times 10^{- 19} J[/tex]
Also, we know that the recoil speed can be calculated by the KInetic energy which is equal to the Energy of the blue-green photon:
[tex]KE_{H} =\frac{1}{2}m_{p}v_{H}[/tex]
where
[tex]v_{H}[/tex] = velocity of Hydrogen atom
[tex]m_{p} = 1.67\times 10^{- 27} kg[/tex] = mass of H-atom
Now,
[tex]KE_{H} =\frac{1}{2}m_{p}(v_{H})^{2}[/tex]
[tex]4.07\times 10^{- 19} =\frac{1}{2}\times 1.67\times 10^{- 27}\times (v_{H})^{2}[/tex]
[tex]v_{H} = \sqrt(4.87\times 10^{8}) = 2.207\times 10^{4} m/s[/tex]
