Answer:
The correct answer is option D.
Explanation:
Using Rydberg's Equation for hydrogen atom:
[tex]\bar{\nu}=\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )[/tex]
Where,
[tex]\bar{\nu}[/tex] = Wave number
[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=5, n_i=3[/tex]
[tex]R_H=1.096776\times 10^7 m^{-1}[/tex]
[tex]\frac{1}{\lambda}=1.096776\times 10^7 m^{-1}\times \left(\frac{1}{3^2}-\frac{1}{5^2} \right )[/tex]
[tex]\frac{1}{\lambda}=1.096776\times 10^7 m^{-1}\times \frac{16}{225}[/tex]
[tex]\lambda =1.2822\times 10^{-6} m=1282.2 nm\approx 1282 nm[/tex]
([tex]1 m= 10^9 nm[/tex])
The wavelength of the photon emitted when the hydrogen atom undergoes a transition from n = 5 to n = 3 is 1282 nm.
The wavelength of the photon is 1282 nm.
From the Rydberg's formula;
1/λ = R (1/n^2final - 1/n^2initial)
R = 1.097 × 10^7 m-1
nfinal = 3
ninitial = 5
Substituting values;
1/λ = 1.096776 x 10^7 m-1 (1/3^2 - 1/5^2)
1/λ = 1.096776 x 10^7 m-1 (1/9 - 1/25)
λ = 1.282 × 10^-6 m
λ = 1282 nm
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