Calculate the wavelength of the line in the Lyman series that results from the transition n=4 to n=1.
Is light absorbed or emitted during this process?

Question

| General Chemistry 1

Accepted Answer

E₄ > E₁ ⇒ E = E₄ – E₁ =

\frac{- 2.1799 \times 10^{- 18}}{\frac{1}{4^{2}} - 1}

= 2.04 x 10^-18J

E =

\frac{hc}{λ}

⇒ λ =

\frac{hc}{E}

= 9.74 x 10^-8 m

E₄ > E₁⇒ emission

Answer

E₄ > E₁ ⇒ E = E₄ – E₁ = $\frac{- 2.1799 \times 10^{- 18}}{\frac{1}{4^{2}} - 1}$ = 2.04 x 10^-18J

E =

\frac{hc}{λ}

⇒ λ =

\frac{hc}{E}

= 9.74 x 10^-8 m

E₄ > E₁⇒ emission

Atomic line spectra | Early Quantum Theory