Chapter 27, Problem 52P

To determine

(a) To Determine:

The wavelength of the second Balmer line.

Answer to Problem 52P

Solution:

The wavelength of the second Balmer line is $486\text{nm}$.

Explanation of Solution

Given:

The transition is from n = 4 to n = 2.

Formula Used:

$\lambda =\frac{(1.24\times {10}^3\text{eV}\text{nm})}{(E_4-E_2)}$

Calculation:

The energy of the 4 state and 2 state can be calculated as

$\begin{array}{l}{E}_4=-\frac{(13.6\text{eV)}}{4^2}=-0.85\text{eV}\\ E_2=-\frac{(13.6\text{eV)}}{2^2}=-3.4\text{eV}\end{array}$

Now by using the formula $\lambda =\frac{(1.24\times {10}^3\text{eV}\text{nm})}{(E_4-E_2)}$, The wavelength of the second Balmer line can be calculated as

$\begin{array}{l}\lambda =\frac{(1.24\times {10}^3\text{eV}\text{nm})}{(E_4-E_2)}\\ \lambda =\frac{(1.24\times {10}^3\text{eV}\text{nm})}{\left[-0.85\text{eV-(-3}\text{.4}\text{eV)}\right]}=486\text{nm}\end{array}$

The wavelength of the second Balmer line is $486\text{nm}$.

To determine

(b) To Determine:

The wavelength of the second Lyman line.

Answer to Problem 52P

Solution:

The wavelength of the second Lyman line is $102\text{nm}$.

Explanation of Solution

Given:

The transition is from n = 3 to n = 1.

Formula Used:

$\lambda =\frac{(1.24\times {10}^3\text{eV}\text{nm})}{(E_3-E_1)}$

Calculation:

The energy of the 4 state and 2 state can be calculated as

$\begin{array}{l}{E}_3=-\frac{(13.6\text{eV)}}{3^2}=-1.5\text{eV}\\ E_1=-\frac{(13.6\text{eV)}}{1^2}=-13.6\text{eV}\end{array}$

Now by using the formula $\lambda =\frac{(1.24\times {10}^3\text{eV}\text{nm})}{(E_3-E_1)}$, The wavelength of the second Lyman line can be calculated as

$\begin{array}{l}\lambda =\frac{(1.24\times {10}^3\text{eV}\text{nm})}{(E_3-E_1)}\\ \lambda =\frac{(1.24\times {10}^3\text{eV}\text{nm})}{\left[-1.5\text{eV-(-13}\text{.6}\text{eV)}\right]}=102\text{nm}\end{array}$

The wavelength of the second Lyman line is $102\text{nm}$.

To determine

(c) To Determine:

The wavelength of the third Balmer line.

Answer to Problem 52P

Solution:

The wavelength of the third Balmer line is $434\text{nm}$.

Explanation of Solution

Given:

The transition is from n = 5 to n = 2.

Formula Used:

$\lambda =\frac{(1.24\times {10}^3\text{eV}\text{nm})}{(E_5-E_2)}$

Calculation:

The energy of the 4 state and 2 state can be calculated as

$\begin{array}{l}{E}_5=-\frac{(13.6\text{eV)}}{5^2}=-0.54\text{eV}\\ E_2=-\frac{(13.6\text{eV)}}{2^2}=-3.4\text{eV}\end{array}$

Now by using the formula $\lambda =\frac{(1.24\times {10}^3\text{eV}\text{nm})}{(E_5-E_2)}$, The wavelength of the second Balmer line can be calculated as

$\begin{array}{l}\lambda =\frac{(1.24\times {10}^3\text{eV}\text{nm})}{(E_5-E_2)}\\ \lambda =\frac{(1.24\times {10}^3\text{eV}\text{nm})}{\left[-0.54\text{eV-(-3}\text{.4}\text{eV)}\right]}=434\text{nm}\end{array}$

The wavelength of the second Balmer line is $434\text{nm}$.