The wave number associated with the Balmer transition for the He^{+} ion (n = 4 to n = 2 ) is given by:

\(\bar{\nu}=\dfrac{1}{\lambda }=RZ^{2}(\dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}})\)

Where, n_{1} = 2 n_{2} = 4

\(\bar{\nu}=\dfrac{1}{\lambda }=R(2)^{2}(\dfrac{1}{4}-\dfrac{1}{16})\\ \\ =4R(\dfrac{4-1}{16})\\ \\ \bar{\nu}=\dfrac{1}{\lambda }=\dfrac{3R}{4}\\ \\ \Rightarrow \lambda =\dfrac{4}{3R}\)

As per the question, the desired transition in the hydrogen spectrum must have the same wavelength as that of He^{+ }spectrum.

\(R(1)^{2}[\dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}}]=\dfrac{3R}{4}\\ \\ \Rightarrow [\dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}}]=\dfrac{3}{4}\: \: \: \: \: …….(1)\)

This equality is true only when the value of n_{1} = 1 and that of n_{2} = 2.

The transition for n_{2} = 2 to n = 1 in the hydrogen spectrum would, therefore, have the same wavelength as the Balmer transition from n = 4 to n = 2 of the He^{+} spectrum.

Answer the following questions:-

**(a)** How many sub-shells are associated with n = 4?

**(b)** How many electrons will be present in the sub-shells having m_{s} value of \(- \dfrac{1}{2}\) for n = 4?

Indicate the number of unpaired electrons in:

**(a)** P

**(b) **Si

**(c)** Cr

**(d)** Fe

**(e)** Kr

The unpaired electrons in Al and Si are present in 3p orbital. Which electrons will experience more effective nuclear charge from the nucleus?

Among the following pairs of orbitals which orbital will experience the larger effective nuclear charge?

**(i) **2s and 3s,

**(ii)** 4d and 4f,

**(iii)** 3d and 3p

The bromine atom possesses 35 electrons. It contains 6 electrons in 2p orbital, 6 electrons in 3p orbital and 5 electrons in 4p orbital. Which of these electron experiences the lowest effective nuclear charge?