Chapter 1, Problem 1C.2E

(a)

Interpretation Introduction

Interpretation:

The wavelength for the transition for helium atom from forth energy level to first energy level has to be determined.

Concept Introduction:

Energy separation between the levels can be calculated using following equation.

    $E_{n_2}-E_{n_1}=\left(n_2^2-n_1^2\right)\frac{h^2}{8m{L}^2}=h\nu$

The frequency $\left(\nu \right)$, wavelength $\left(\lambda \right)$ and velocity of light $\left(c\right)$ are related by following equation.

    $\lambda =\frac{c}{\nu }$

Answer to Problem 1C.2E

The wavelength for the transition for helium atom is $2.192nm.$

Explanation of Solution

Transition is taking place from forth energy level to first energy level.

The length of the box is given as $100pm.$

The energy difference for the transition is given below.

    $\begin{array}{c}{E}_{n_2}-E_{n_1}=\left(n_2^2-n_1^2\right)\frac{h^2}{8m{L}^2}=h\nu \\ \\ E_4-E_1=\frac{15h^2}{8m{L}^2}=h\nu \\ \\ \frac{15h}{8m{L}^2}=\nu \end{array}$

Therefore the wavelength can be calculated as follows,

    $\begin{array}{c}\lambda =\frac{c}{\nu }\\ \\ \lambda =\frac{8mc{L}^2}{15h}\\ \\ =\frac{8\left(9.10939\times {10}^{-31}kg\right)\times \left(2.988\times {10}^{8}m/s\right){\left({10}^{-10}m\right)}^2}{15\times \left(6.626\times {10}^{-34}J.s\right)}\\ \\ =2.192\times {10}^{-9}m\\ \\ =2.192nm.\end{array}$

Wavelength is $2.192nm.$

(b)

Interpretation Introduction

Interpretation:

The wavelength for the transition for hydrogen atom from forth energy level to second energy level has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 1C.2E

The wavelength for the transition for helium atom is $2.74nm.$

Explanation of Solution

Transition is taking place from forth energy level to second energy level.

The length of the box is given as $100pm.$

The energy difference for the transition is given below.

    $\begin{array}{c}{E}_{n_2}-E_{n_1}=\left(n_2^2-n_1^2\right)\frac{h^2}{8m{L}^2}=h\nu \\ \\ E_4-E_2=\frac{12h^2}{8m{L}^2}=h\nu \\ \\ \frac{12h}{8m{L}^2}=\nu \end{array}$

Therefore the wavelength can be calculated as follows,

    $\begin{array}{c}\lambda =\frac{c}{\nu }\\ \\ \lambda =\frac{8mc{L}^2}{12h}\\ \\ =\frac{8\left(9.10939\times {10}^{-31}kg\right)\times \left(2.988\times {10}^{8}m/s\right){\left({10}^{-10}m\right)}^2}{12\times \left(6.626\times {10}^{-34}J.s\right)}\\ \\ =2.74\times {10}^{-9}m\\ \\ =2.74nm.\end{array}$

Wavelength is $2.74nm.$