The \(n_{i} = 4\; to\; n_{f}= 2\) transition results in a spectral line of the Balmer series. The energy involved in this transition can be calculated using the following expression:

\(=2.18\times 10^{-18}[\dfrac{1}{n_{i}^{2}}-\dfrac{1}{n_{f}^{2}}]\)

Substituting these values in the expression for E:

\(E=2.18\times 10^{-18}[\dfrac{1}{4^{2}}-\dfrac{1}{2^{2}}]\\ \\ =2.18\times 10^{-18}[\dfrac{1-4}{16}]\\ \\ =2.18\times 10^{-18}\times (-\dfrac{3}{16})\\ \\ E=-(4.0875\times 10^{-19}J)\)

Here, the -ve sign denotes the emitted energy.

Wavelength of the emitted light \((\lambda )=\dfrac{hc}{E}\)

(Since \(E=\dfrac{hc}{\lambda })\)

Substituting these values in the expression for \({\lambda }\)

\(\lambda =\dfrac{(6.626\times 10^{-34})(3\times 10^{8})}{(4.0875\times 10^{-19})}\)

\( =4.8631\times 10^{-7}\, m\\ \\ \)

\( \lambda =486.31\times 10^{-9}\, m\\ \\ \)

\( =486\, nm\)

Answered by Pragya Singh | 1 year agoAnswer 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?