Chapter 28, Problem 60GP

To determine

The Z and the required element.

Answer to Problem 60GP

The value of $Z=4$ and element is beryllium.

Explanation of Solution

Given info:

The spectrum of the unknown element shows the series of jines with one out of every four matching a line from the Lyman series of hydrogen.

Assume that the element is the ion with the Z protons.

Calculation

Consider the energy states of hydrogen atom.

Consider the wavelength of the transition from the bohr formula is given by,

  $\frac{1}{\lambda }\alpha Z^2\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]$

As the atomic number of the hydrogen is $Z=1$ .

Consider the lower energy state of the Lyman series is $n_1$ .

Then, the wavelength of the Lyman series in hydrogen is given by,

  $\frac{1}{\lambda }\alpha \left[\frac{1}{1}-\frac{1}{n^2{}_2}\right]$ …… (1)

Let the atomic number of the unknown atom is Z and the wavelength out of every four transitions between n to n coincides with the Lyman series of hydrogen atom.

Then, the wavelength of the Lyman series in hydrogen is given by,

  $\frac{1}{\lambda }\alpha Z^2\left[\frac{1}{n_{1x}^2}-\frac{1}{n^2{}_{2x}}\right]$ …… (2)

Consider the wavelength of both the series coincide. Then, from equation 1 and 2

  $\left[\frac{1}{1}-\frac{1}{n^2{}_2}\right]=Z^2\left[\frac{1}{n_{1x}^2}-\frac{1}{n^2{}_{2x}}\right]$

The above case is possible when the $n_{1x}=4$ and $Z=4$ .

Thus, the element is beryllium.

Conclusion:

Therefore, the value of $Z=4$ and element is beryllium.