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

`n_1=1.459`

Step-by-step explanation:

Hello,

In this case, the Rydberd formula is:

`(1)/(\lambda)=RZ^2((1)/(n_1^2)-(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:

`(1)/(n_1^2)-(1)/(n_2^2) =(1)/(\lambda RZ^2)\\(1)/(n_1^2)=(1)/(\lambda RZ^2)+(1)/(n_2^2)\\(1)/(n_1^2)=(1)/(4.141x10^(-7)m*1.097x10^7m^(-1)*1^2)+(1)/(2^2) \\(1)/(n_1^2)=0.47\\n_1=\sqrt{(1)/(0.47)} \\n_1=1.459`

Best regards.

Answer 2

1

Step-by-step explanation:

Using the Rydberg formula as:

`\frac {1}{\lambda}=R_H* Z^2* (\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* 10^(-9)}=1.097* 10^7* 1^2* (\frac {1}{n_(1)^2}-\frac {1}{2^2})`

Solving for n₁ , we get

n₁ ≅ 1