Chapter 8, Problem 8.4P

(a)

Interpretation Introduction

Interpretation:

The wavelength of the next line in the given series of lines of spectrum of atomic hydrogen has to be identified.

Explanation of Solution

All the lines in the hydrogen spectrum suits the Rydberg formula.

    $\begin{array}{l}\frac{\text{1}}{\text{λ}}\text{=}{\text{R}}_{\text{H}}\left(\frac{\text{1}}{{\text{n}}_{\text{1}}^{\text{2}}}\text{-}\frac{\text{1}}{{\text{n}}_{\text{2}}^{\text{2}}}\right)\\ \\ \text{Where}\text{λ}\text{is}\text{the}\text{wavelength}\text{of}\text{the}\text{emitted}\text{photon}\\ \text{R is Rydberg’s constant}\\ \text{Z is the atomic number }\left(\text{1 for H, 2 for He}\right)\\ {\text{n}}_{\text{1}}{\text{ and n}}_{\text{2}}\text{ are the principal quantum numbers for the initial and final states}\end{array}$

Find ${\text{n}}_{\text{1}}$ from the value of  ${\text{λ}}_{\text{max}}$, which arises from the transition of ${\text{n}}_{\text{1}}+1\underset{}{\overset{}{\to }}{\text{n}}_{\text{1}}$.

    $\begin{array}{l}\frac{\text{1}}{{\text{λ}}_{\text{max}}{\text{R}}_{\text{H}}}\text{=}\frac{\text{1}}{{\text{n}}_{\text{1}}^{\text{2}}}\text{-}{\left(\frac{\text{1}}{{\text{n}}_{\text{1}}\text{+}\text{1}}\right)}^{\text{2}}\text{=}\frac{{\text{2n}}_{\text{1}}\text{+}\text{1}}{{\text{n}}_{\text{1}}^{\text{2}}{{\text{(n}}_{\text{1}}\text{+}\text{1)}}^{\text{2}}}\\ \\ {\text{λ}}_{\text{max}}{\text{R}}_{\text{H}}\text{=}\frac{{\text{n}}_{\text{1}}^{\text{2}}{{\text{(n}}_{\text{1}}\text{+}\text{1)}}^{\text{2}}}{{\text{2n}}_{\text{1}}\text{+}\text{1}}\text{=}\text{(656}\text{.46}\text{×}{\text{10}}^{\text{-9}}\text{m)}\text{×}\text{(109677×}{\text{10}}^{\text{2}}{\text{m}}^{\text{-1}}\text{)}\\ \\ \text{=}\text{7}\text{.20}\\ \text{hence}{\text{n}}_{\text{1}}\text{=}\text{2,}\text{as}\text{determined}\text{by}\text{trial}\text{and}\text{error}\text{substitution}\text{.}\text{Therefore,}\\ \text{the}\text{transitions}\text{are}\text{given}\text{as,}\\ \\ \text{υ}\text{=}\frac{\text{1}}{\text{λ}}\text{=}\text{(109677}{\text{cm}}^{\text{-1}}\text{)}\text{×}\left(\frac{\text{1}}{\text{4}}\text{-}\frac{\text{1}}{{\text{n}}_{\text{2}}^{\text{2}}}\right){\text{n}}_{\text{2}}\text{=}\text{3,4,5,6}\\ \\ \text{Next}\text{line}\text{has}{\text{n}}_{\text{2}}\text{=}\text{7}\\ \\ \text{υ}\text{=}\frac{\text{1}}{\text{λ}}\text{=}\text{(109677}{\text{cm}}^{\text{-1}}\text{)}\text{×}\left(\frac{\text{1}}{\text{4}}\text{-}\frac{\text{1}}{\text{49}}\right)\text{=}\text{397}\text{.13nm}\end{array}$

The energy needed to ionize the atom is obtained by allowing ${\text{n}}_2\underset{}{\overset{}{\to }}\infty$

    $\begin{array}{l}\text{υ}\text{=}\frac{\text{1}}{{\text{λ}}_{\infty }}\text{=}\text{(109677}{\text{cm}}^{\text{-1}}\text{)}\text{×}\left(\frac{\text{1}}{\text{4}}\text{-}\text{0}\right)\text{=}{\text{27419cm}}^{\text{-1}}\\ \\ \text{υ}\text{=}\text{3}\text{.40ev}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The ionization energy of hydrogen atom when it is in lower state of transition has to be identified.

Explanation of Solution

The energy needed to ionize the atom is obtained by allowing ${\text{n}}_2\underset{}{\overset{}{\to }}\infty$

    $\begin{array}{l}\text{υ}\text{=}\frac{\text{1}}{{\text{λ}}_{\infty }}\text{=}\text{(109677}{\text{cm}}^{\text{-1}}\text{)}\text{×}\left(\frac{\text{1}}{\text{4}}\text{-}\text{0}\right)\text{=}{\text{27419cm}}^{\text{-1}}\\ \\ {\text{1cm}}^{\text{-1}}\text{=}\text{1}\text{.2398}{\text{×10}}^{\text{-4}}\text{ev}\\ \\ \text{υ}\text{=}\text{3}\text{.40ev}\end{array}$