The wavelength of a spectral line in one of the spectral series of atomic hydrogen is 12825 mathringA. What is the wavelength ( in mathringA) of the series limit?

Question

The wavelength of a spectral line in one of the spectral series of atomic hydrogen is 12825 mathringA. What is the wavelength ( in mathringA) of the series limit? (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Δ E=(1240/1282.5) eV =0.96 eV This is the difference in levels 5 and 3 of atomic hydrogen. Now series limit means 3 arrow ∞ (1/λ)=(912)-1((1/n12)) ⇒ (1/λ)=(912)-1((1/9)) ⇒ λ=8208 mathringA

Q. The wavelength of a spectral line in one of the spectral series of atomic hydrogen is $12825 \overset{˚}{A}$ . What is the wavelength $($ in $\overset{˚}{A})$ of the series limit?

$Δ E = \frac{1240}{1282.5} e V = 0.96 e V$
This is the difference in levels $5$ and $3$ of atomic hydrogen.
Now series limit means $3 \to \infty$
$\frac{1}{λ} = (912)^{- 1} (\frac{1}{n _{1}^{2}})$
$\Rightarrow \frac{1}{λ} = (912)^{- 1} (\frac{1}{9})$
$\Rightarrow λ = 8208 \overset{˚}{A}$

Q. The wavelength of a spectral line in one of the spectral series of atomic hydrogen is 12825A˚. What is the wavelength ( in A˚) of the series limit?

ΔE=1282.51240​eV=0.96eV This is the difference in levels 5 and 3 of atomic hydrogen. Now series limit means 3→∞ λ1​=(912)−1(n12​1​) ⇒λ1​=(912)−1(91​) ⇒λ=8208A˚

Q. The wavelength of a spectral line in one of the spectral series of atomic hydrogen is $12825 \overset{˚}{A}$ . What is the wavelength $($ in $\overset{˚}{A})$ of the series limit?

$Δ E = \frac{1240}{1282.5} e V = 0.96 e V$
This is the difference in levels $5$ and $3$ of atomic hydrogen.
Now series limit means $3 \to \infty$
$\frac{1}{λ} = (912)^{- 1} (\frac{1}{n _{1}^{2}})$
$\Rightarrow \frac{1}{λ} = (912)^{- 1} (\frac{1}{9})$
$\Rightarrow λ = 8208 \overset{˚}{A}$