Chapter 3, Problem 3.18P

To determine

The wavelength frequency, and energy (electron volts) for the second and third lines in the Lyman Series.

Answer to Problem 3.18P

Second Line λ = 1.023 x 10-7 m

E = 12.11 eV

Third Line $\begin{array}{l}\lambda =9.7\times {10}^{-8}\text{m}\\ E=12.78\text{eV}\end{array}$

Explanation of Solution

Formula used:

The formulas $\frac{1}{\lambda }=R(\frac{1}{n_1^2}-\frac{1}{n_2^2})$ and $c=f\lambda$ is used to compute the wavelength frequency and energy.

Calculation:

To solve the second line:

  $R=1.097\times {10}^7{m}^{-1}$

  $c=3\times {10}^{8}m/sec$

n₁= 1

n₂ = 3

Substituting these values,

  $\begin{array}{l}\frac{1}{\lambda }=R(\frac{1}{n_1^2}-\frac{1}{n_2^2})=\text{1}.0\text{97}\times \text{1}{0}^{\text{7}}\left[\left(\text{1}/{\text{1}}^{\text{2}}\right)-\left(\text{1}/{\text{3}}^{\text{2}}\right)\right]=\text{9}.\text{76}\times \text{1}{0}^{\text{6}}{\text{m}}^{-\text{1}}\\ \text{1}.0\text{97}\times \text{1}{0}^{\text{7}}\left[\left(\text{1}/{\text{1}}^{\text{2}}\right)-\left(\text{1}/{\text{3}}^{\text{2}}\right)\right]=\text{9}.\text{76}\times \text{1}{0}^{\text{6}}{\text{m}}^{-\text{1}}\end{array}$

  $\lambda =1.023\times {10}^{-7}m$

To determine the energy, the frequency must be found.

  $f=\frac{c}{\lambda }=\frac{3\times {10}^{8}m/sec}{1.023\times {10}^{-7}m}=2.93\times {10}^{15}{\sec }^{-1}$ , which is the frequency of second line.

  $h=6.625\times {10}^{-34}J\sec$

  $E=hf=6.625\times {10}^{-34}J.\sec \times 2.93\times {10}^{15}{\sec }^{-1}=1.94\times {10}^{-18}J$

  $=1.94\times {10}^{-18}J\times (\frac{1eV}{1.6\times {10}^{-19}J})=12.11eV$

To solve the third line:

  $R=1.097\times {10}^7{m}^{-1}$

  $c=3\times {10}^{8}m/sec$

n₁= 1

n₂ = 4

Substituting these values,

  $\frac{1}{\lambda }=R(\frac{1}{n_1^2}-\frac{1}{n_2^2})=1.097\times {10}^7(\frac{1}{1_1^2}-\frac{1}{4_2^2})=10.28\times {10}^6{m}^{-1}$

  $\lambda =9.7\times {10}^{-8}m$

To determine the energy, the frequency must be found.

  $f=\frac{c}{\lambda }=\frac{3\times {10}^{8}m/sec}{9.7\times {10}^{-8}m}=3.085\times {10}^{15}{\sec }^{-1}$ , which is the frequency of second line.

  $h=6.625\times {10}^{-34}J\sec$

  $E=hf=6.625\times {10}^{-34}J.\sec \times 3.085\times {10}^{15}{\sec }^{-1}=2.044\times {10}^{-18}J$

  $=2.044\times {10}^{-18}J\times (\frac{1eV}{1.6\times {10}^{-19}J})=12.78eV$

Conclusion:

The wavelength of the second line λ is $1.023\times {10}^{-7}m$ and the energy E is 12.11 eV. The wavelength of the third line λ is $9.7\times {10}^{-8}m$ and the energy E is 12.78eV.