Question

Using the Rydberg formula, calculate the initial energy level when an electron in a hydrogen atom transitions into n= 2 and emits a photon at 410.1 nm. Note: the Rydberg constant = 1.097 x 107 m–1 .

Answer 1

Answer:

$n_1=1.459$

Explanation:

Hello,

In this case, the Rydberd formula is:

$\frac{1}{\lambda}=RZ^2(\frac{1}{n_1^2}-\frac{1}{n_2^2} )$

Whereas $\lambda$ is wavelength of photon,  $R$ the Rydberg's constant,  $Z$ the atomic number of the atom, in this case 1 as it is hydrogen,  $n_1$ is the initial level and $n_2$ is the final energy level .

In such a way, solving for $n_1$ one obtains:

$\frac{1}{n_1^2}-\frac{1}{n_2^2} =\frac{1}{\lambda RZ^2}\\\frac{1}{n_1^2}=\frac{1}{\lambda RZ^2}+\frac{1}{n_2^2}\\\frac{1}{n_1^2}=\frac{1}{4.141x10^{-7}m*1.097x10^7m^{-1}*1^2}+\frac{1}{2^2} \\\frac{1}{n_1^2}=0.47\\n_1=\sqrt{\frac{1}{0.47}} \\n_1=1.459$

Best regards.

Answer 2

Answer:

1

Explanation:

Using the Rydberg formula as:

$\frac {1}{\lambda}=R_H\times Z^2\times (\frac {1}{n_{1}^2}-\frac {1}{n_{2}^2})$

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:

$\frac {1}{410.1\times 10^{-9}}=1.097\times 10^7\times 1^2\times (\frac {1}{n_{1}^2}-\frac {1}{2^2})$

Solving for n₁ , we get

n₁ ≅ 1