Respuesta :
Answer:
[tex]\Delta E=E_{final}-E_{initial}[/tex]
[tex]\Delta E=-1312[\frac{1}{(n_f^2)}-\frac {1}{(n_i^2 )}]KJ mol^{-1}[/tex]
[tex]\Delta E=-1312[\frac{1}{3^2)}-\frac {1}{(1^2 )}]KJ mol^{-1}[/tex]
[tex]\Delta E=-1312[\frac{1}{(9)}-\frac {1}{(1 )}]KJ mol^{-1}[/tex]
[tex]\Delta E=-1312[0.111-1]KJ mol^{-1}[/tex]
[tex]\Delta E=1166 KJ mol^{-1}[/tex]
[tex]\frac{=1166,000 \mathrm{J}}{6.022 \times 10^{23} \text { photons }}[/tex]
[tex]=193623 \times 10^{-23} \frac {J}{photon}[/tex]
[tex]\Delta E=1.93623 \times 10^{-18} \frac {J}{photon}[/tex]
[tex]\Delta E=\frac {h\times c}{\lambda} \\\\=\frac {(6.626\times 10^{-34} J s \times 3 \times 10^8 ms^{-1})}{\lambda}[/tex]
h is planck's constant
c is the speed of light
λ is the wavelength of light
[tex]\lambda =\frac {h\times c}{\Delta E}\\\\=\frac {(6.626\times10^{-34} J s\times3 \times 10^8 ms^{-1})}{(1.93623\times10^{-18} J/photon)}[/tex]
Wavelength
[tex]\lambda =10.3 \times 10^{-8} m \times \frac {(10^9 nm)}{1m} =103 nm (Answer)[/tex]
Thus, the wavelength of light associated with the transition from n=1 to n=3 in the hydrogen atom is 103 nm.
We have that for the Question "calculate the wavelength of light associated with the transition from n=1 to n=3 in the hydrogen atom?" it can be said that the wavelength of light associated with the transition from n=1 to n=3 in the hydrogen atom is
[tex]\lambda=103nm[/tex]
From the question we are told
calculate the wavelength of light associated with the transition from n=1 to n=3 in the hydrogen atom?
Generally the equation for the wavelength is mathematically given as
[tex]1/\lambda = -R* (\frac{1}{nf^2} - \frac{1}{ni^2}\\\\\lambda = \frac{1}{-( 1.097*10^7)* (\frac{1}{3^2} - \frac{1}{1^2}}[/tex]
[tex]\lambda=103nm[/tex]
Therefore
the wavelength of light associated with the transition from n=1 to n=3 in the hydrogen atom is
[tex]\lambda=103nm[/tex]
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