Respuesta :
Answer:
The wavelength of the light emitted by a hydrogen atom for the given transition is 2166 nm.
Explanation:
The energy of nth energy levels of the H atom is given as:
[tex]E_n = -2.18 \times 10^{-18} \times \frac{1}{n^2} J[/tex]
Energy of the seventh energy level = [tex]E_7[/tex]
[tex]E_7=-2.18 \times 10^{-18} \times \frac{1}{7^2} J[/tex]
[tex]E_7=-2.18 \times 10^{-18} \times \frac{1}{7^2} J=-4.4490\times 10^{-20} J[/tex]
Energy of the seventh energy level = [tex]E_4[/tex]
[tex]E_4=-2.18 \times 10^{-18} \times \frac{1}{4^2} J[/tex]
[tex]E_4=-2.18 \times 10^{-18} \times \frac{1}{16} J=-1.3625\times 10^{-19} J[/tex]
Energy of the light emitted will be equal to the energy difference of the both levels.
[tex]E=E_7-E_4=-4.4490\times 10^{-20} J-(-1.3625\times 10^{-19} J)[/tex]
[tex]E=9.176\times 10^{-20} J[/tex]
Wavelength corresponding to energy E can be calculated by using Planck's equation:
[tex]E=\frac{hc}{\lambda }[/tex]
[tex]\lambda =\frac{hc}{E}=\frac{6.626\times 10^{-34} Js\times 3\times 10^8 m/s}{9.176\times 10^{-20} J}=2.166\times 10^{-6} m=2166 nm[/tex]
The wavelength of the light emitted by a hydrogen atom for the given transition is 2166 nm.
Answer:
The wavelength of the light emitted by electron fall will be 216.62 nm.
Explanation:
The change in the energy state of the matter can be obtained from the principal quantum number of the shells.
Energy of 4th shell - Energy of 7th shell = [tex]\rm \Delta[/tex]E
[tex]\Delta[/tex]E = [tex]\rm 2.18\;\times\;10^-^1^8\;J\;\left ( \frac{1}{n_f^2} \;-\;\frac{1}{n_i^2} \right )[/tex]
[tex]\rm \Delta\;E\;=\;-2.18\;\times\;10^-^1^8\;\left (\frac{1}{4^2} \;-\;\frac{1}{7^2} \right )[/tex]
[tex]\rm \Delta\;E\;=\;-2.18\;\times\;10^-^1^8\;\times\;0.0421[/tex]
[tex]\rm \Delta\;E\;=\;0.0917\;\times\;10^-^1^8[/tex] J
The wavelength can be calculated as:
[tex]\rm \Delta\;E\;=\;\frac{hc}{\lambda}[/tex]
where, h is Plank's constant,
c is speed of light
[tex]\lambda[/tex] is the wavelength
[tex]\rm 0.0917\;\times\;10^-^1^8\;J\;=\;\frac{6.626\;\times\;10^-^3^4\;J.s\;\times\;2.998\;\times\;10^8\;m/s}{\lambda}[/tex]
[tex]\rm\lambda\;=\;216.62nm[/tex].
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