Answer: The wavelength of light is [tex]3.97\times 10^2nm[/tex]
Explanation:
To calculate the wavelength of light, we use Rydberg's Equation:
[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]1.097\times 10^7m^{-1}[/tex]
[tex]n_f[/tex] = Higher energy level = 7
[tex]n_i[/tex]= Lower energy level = 2
Putting the values in above equation, we get:
[tex]\frac{1}{\lambda }=1.097\times 10^7m^{-1}\left(\frac{1}{2^2}-\frac{1}{7^2} \right )\\\\\lambda =\frac{1}{0.2518\times 10^7m^{-1}}=3.97\times 10^{-7}m[/tex]
Converting this into nanometers, we use the conversion factor:
[tex]1m=10^9nm[/tex]
So, [tex]3.97\times 10^{-7}m\times (\frac{10^9nm}{1m})=3.97\times 10^2nm[/tex]
Hence, the wavelength of light is [tex]3.97\times 10^2nm[/tex]
The wavelength of the light absorbed when an electron in a hydrogen atom transitions from an orbital in which n = 2 to an orbital in which n = 7 is 397 nm.
An electron in a hydrogen atom makes a transition from an orbital in which n = 2 (ni) to an orbital in which n = 7 (nf).
We can calculate the wavelength (λ) of the light absorbed using Rydberg's equation.
[tex]\frac{1}{\lambda } = R_{H}(\frac{1}{n_i^{2} }-\frac{1}{n_f^{2} } )[/tex]
where,
[tex]\frac{1}{\lambda } = R_{H}(\frac{1}{n_i^{2} }-\frac{1}{n_f^{2} } ) = (1.097 \times 10^{7}m^{-1} )(\frac{1}{2^{2} }-\frac{1}{7^{2} } )\\\\\lambda = 3.97 \times 10^{-7} m \times \frac{1nm}{10^{-9}m } = 397nm[/tex]
The wavelength of the light absorbed when an electron in a hydrogen atom transitions from an orbital in which n = 2 to an orbital in which n = 7 is 397 nm.
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