Respuesta :
Answer:
[tex]\lambda=103\ nm[/tex]
Explanation:
[tex]E_n=-2.179\times 10^{-18}\times \frac{1}{n^2}\ Joules[/tex]
For transitions:
[tex]Energy\ Difference,\ \Delta E= E_f-E_i =-2.179\times 10^{-18}(\frac{1}{n_f^2}-\frac{1}{n_i^2})\ J=2.179\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J[/tex]
[tex]\Delta E=2.179\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J[/tex]
Also, [tex]\Delta E=\frac {h\times c}{\lambda}[/tex]
Where,
h is Plank's constant having value [tex]6.626\times 10^{-34}\ Js[/tex]
c is the speed of light having value [tex]3\times 10^8\ m/s[/tex]
So,
[tex]\frac {h\times c}{\lambda}=2.179\times 10^{-18}(|\frac{1}{n_i^2} - \dfrac{1}{n_f^2}|)\ J[/tex]
[tex]\lambda=\frac {6.626\times 10^{-34}\times 3\times 10^8}{{2.179\times 10^{-18}}\times (|\frac{1}{n_i^2} - \dfrac{1}{n_f^2}|)}\ m[/tex]
So,
[tex]\lambda=\frac {6.626\times 10^{-34}\times 3\times 10^8}{{2.179\times 10^{-18}}\times (|\frac{1}{n_i^2} - \dfrac{1}{n_f^2}|)}\ m[/tex]
Given, [tex]n_i=3\ and\ n_f=1[/tex]
[tex]\lambda=\frac{6.626\times 10^{-34}\times 3\times 10^8}{{2.179\times 10^{-18}}\times |(\frac{1}{3^2} - \dfrac{1}{1^2})}|\ m[/tex]
[tex]\lambda=\frac{10^{-26}\times \:19.878}{10^{-18}\times \:2.179|\left(\frac{1}{9}-\frac{1}{1}\right)|}\ m[/tex]
[tex]\lambda=1.03\times 10^{-7}\ m[/tex]
1 m = 10⁻⁹ nm
[tex]\lambda=103\ nm[/tex]
The wavelength of the photon emitted during transition from the [tex]n = 3[/tex] state to the [tex]n = 1[/tex] state in the H atom is 102.55 Nanometer.
Given the data in the equation;
[tex]n_1 = 1[/tex]
[tex]n_2 = 3[/tex]
Wavelength; [tex]\lambda = \ ?[/tex]
To determine the wavelength of the photon, we use the Rydberg Equation:
[tex]\frac{1}{\lambda} = R[ \frac{1}{n_1^2} - \frac{1}{n_2^2} ][/tex]
Where R is the Rydberg constant ( [tex]1.097 * 10^7 m^{-1}[/tex] )
We substitute our values into the equation and solve for the Wavelength
[tex]\frac{1}{\lambda} = (1.097 * 10^7 m^{-1})[ \frac{1}{(1)^2} - \frac{1}{(3)^2} ]\\\\\frac{1}{\lambda} = (1.097 * 10^7 m^{-1}) 0.8888\\\\\frac{1}{\lambda} = 9.751 * 10^6 m^{-1}\\\\\lambda = \frac{1}{9.751 * 10^6 m^{-1}} \\\\\lambda = 1.0255 * 10^{-7}m\\\\\lambda = 102.55nm[/tex]
Therefore, the wavelength of the photon emitted during transition from the [tex]n = 3[/tex] state to the [tex]n = 1[/tex] state in the H atom is 102.55 Nanometer.
Learn more: https://brainly.com/question/5305512
