What is the energy (in J) of a photon that is ejected from a hydrogen atom when an electron relaxes from the 6th to the 3rd energy level ?

Respuesta :

Answer: [tex]1.8\times 10^{-19}J[/tex]

Explanation:

Using Rydberg's Equation for hydrogen atom:

[tex]\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )[/tex]

Where,

[tex]\lambda[/tex] = Wavelength of radiation

[tex]R_H[/tex] = Rydberg's Constant

[tex]n_f[/tex] = Higher energy level

[tex]n_i[/tex]= Lower energy level

We have:

[tex]n_f=6, n_i=3[/tex]

[tex]R_H=1.09\times 10^7 m^{-1}[/tex]

[tex]\frac{1}{\lambda}=1.09\times 10^7 m^{-1}\times \left(\frac{1}{3^2}-\frac{1}{6^2} \right )[/tex]

[tex]\frac{1}{\lambda}=1.09\times 10^7 m^{-1}\times \frac{3}{36}[/tex]

[tex]\frac{1}{\lambda}=0.0908\times 10^{7}m^{-1}[/tex]

[tex]\lambda=11.0\times 10^{-7} m[/tex]

The relation between energy and wavelength of light is given by Planck's equation, which is:

[tex]E=\frac{hc}{\lambda}[/tex]

where,

E = energy of the light

h = Planck's constant

c = speed of light

[tex]\lambda[/tex] = wavelength of light

[tex]E=\frac{6.6\times 10^{-34}\times 3\times 10^8}{11.0\times 10^{-7}m}[/tex]

[tex]E=1.8\times 10^{-19}J[/tex]

Thus energy of a photon that is ejected from a hydrogen atom when an electron relaxes from the 6th to the 3rd energy level is [tex]1.8\times 10^{-19}J}[/tex]

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