Answer:
1
Explanation:
Using the Rydberg formula as:
[tex]\frac {1}{\lambda}=R_H\times Z^2\times (\frac {1}{n_{1}^2}-\frac {1}{n_{2}^2})[/tex]
where,
λ is wavelength of photon
R = Rydberg's constant (1.097 × 10⁷ m⁻¹)
Z = atomic number of atom
n₁ is the initial final level and n₂ is the final energy level
For Hydrogen atom, Z= 1
n₂ = 2
Wavelength = 410.1 nm
Also,
1 nm = 10⁻⁹ m
So,
Wavelength = 410.1 × 10⁻⁹ m
Applying in the formula as:
[tex]\frac {1}{410.1\times 10^{-9}}=1.097\times 10^7\times 1^2\times (\frac {1}{n_{1}^2}-\frac {1}{2^2})[/tex]
Solving for n₁ , we get
n₁ ≅ 1
Answer:
[tex]n_1=1.459[/tex]
Explanation:
Hello,
In this case, the Rydberd formula is:
[tex]\frac{1}{\lambda}=RZ^2(\frac{1}{n_1^2}-\frac{1}{n_2^2} )[/tex]
Whereas [tex]\lambda[/tex] is wavelength of photon, [tex]R[/tex] the Rydberg's constant, [tex]Z[/tex] the atomic number of the atom, in this case 1 as it is hydrogen, [tex]n_1[/tex] is the initial level and [tex]n_2[/tex] is the final energy level .
In such a way, solving for [tex]n_1[/tex] one obtains:
[tex]\frac{1}{n_1^2}-\frac{1}{n_2^2} =\frac{1}{\lambda RZ^2}\\\frac{1}{n_1^2}=\frac{1}{\lambda RZ^2}+\frac{1}{n_2^2}\\\frac{1}{n_1^2}=\frac{1}{4.141x10^{-7}m*1.097x10^7m^{-1}*1^2}+\frac{1}{2^2} \\\frac{1}{n_1^2}=0.47\\n_1=\sqrt{\frac{1}{0.47}} \\n_1=1.459[/tex]
Best regards.
