Question

If the shorts series limit of the Balmer series for hydrogen is 3646 \[{{A}^{0}}\], find the atomic number of the element which gives X-ray wavelength down to 1.0 \[{{A}^{0}}\]. Identify the element.

Answer 1

Hint: Balmer series: This series is present due to the transition of the electrons from different outer orbit to the second orbit. Therefore the wavelength of the Balmer series is given by,

$\dfrac{1}{\lambda }=R\left( \dfrac{1}{{{2}^{2}}}-\dfrac{1}{{{n}^{2}}} \right){{\left( Z \right)}^{2}};where,n=3,4,5....$

For n= 3 and 4, the wavelength for Hydrogen atom(z=1) is 6563${{A}^{0}}$ and 4868${{A}^{0}}$ respectively. As the value of n increases wavelength decreases. It means frequency will increase.

Complete step by step solution:

Here the question is talking about the shortest series. The shortest series of Balmer must have maximum energy. Means if the wavelength is minimum then energy must be maximum.

We will get maximum energy when the energy level will be high, the gap will be more.
For the Balmer series, n is 2. So our n will initiate from infinity to 2

i.e. ${{n}_{f}}=\infty \And {{n}_{i}}=2$

let’s find the value of R

$\dfrac{1}{\lambda }=R\left( \dfrac{1}{{{2}^{2}}}-\dfrac{1}{{{n}^{2}}} \right){{1}^{2}}$
$\dfrac{1}{\lambda }=R\left( \dfrac{1}{4}-\dfrac{1}{{{\infty }^{2}}} \right)$
$\dfrac{1}{\lambda }=\dfrac{R}{4}$
$R=\dfrac{4}{\lambda }$

Given wavelength is 3646\[{{A}^{0}}\]

Therefore, \[R=\dfrac{4}{3646}{{A}^{{{0}^{-1}}}}\]

Now the atomic number of the element which gives X-ray wavelength down to 1.0\[{{A}^{0}}\] i.e. for n=1:

$\dfrac{1}{\lambda }=R\left( \dfrac{1}{{{1}^{2}}}-\dfrac{1}{{{n}^{2}}} \right)$
$\dfrac{1}{{{\lambda }^{'}}}=R\left( \dfrac{1}{1}-\dfrac{1}{{{\infty }^{2}}} \right){{\left( z-1 \right)}^{2}}$
${{\left( z-1 \right)}^{2}}=\dfrac{1}{\lambda 'R}$
${{\left( z-1 \right)}^{2}}=\dfrac{1}{1\times \dfrac{4}{3646}}$

$\begin{align}
  & z-1=\sqrt{911.5} \\
 & z-1=30.19 \\
 & z=31.2\approx 31 \\
 & z=31 \\
\end{align}$

We know that the atomic number of gallium is 31

The atomic number of the element which gives X-ray wavelength down to 1.0\[{{A}^{0}}\] is gallium

Note: For Balmer series value of n starts from 2. In the first case an element was known that is hydrogen and an atomic number of hydrogen is 1. But later, in question, it is given that the value of an atomic number is decreased by one there in formula z is replaced by z-1. Always read the question carefully and try to imagine what solution should be and for what they are asking.