The radius of the n th orbit of hydrogen-like particles is given by,

\(r=\dfrac{0.529n^{2}}{Z}Å\)

\(r=\dfrac{5.29n^{2}}{Z}pm\)

For radius (r_{1}) = 1.3225 nm

\(=1.32225\times 10^{-9}m\\ \\ =1322.25\times 10^{-12}m\\ \\ =1322.25\, pmn_{1}^{2}\)

\( =\dfrac{r_{1}Z}{52.9}\\ \\ n_{1}^{2}=\dfrac{1322.25Z}{52.9}\)

Similarly,

\(n_{2}^{2}=\dfrac{211.6Z}{52.9}\\ \\ \dfrac{n_{1}^{2}}{n_{2}^{2}}=\dfrac{1322.5}{211.6}\\ \\ \dfrac{n_{1}^{2}}{n_{2}^{2}}=6.25\\ \\ \dfrac{n_{1}}{n_{2}}=2.5\\ \\ \dfrac{n_{1}}{n_{2}}=\dfrac{25}{10}=\dfrac{5}{2}\)

⇒ n1 = 5 and n2 = 2

Therefore, the electron transition is from the 5th orbit to the 2nd orbit and it, therefore, corresponds to the Balmer series.

The wave number \(\bar{\nu}\) of the transition is:

\(1.097\times 10^{7}(\dfrac{1}{2^{2}}-\dfrac{1}{5^{2}})m^{-1}\\ \\ =1.097\times 10^{7}m^{-1}(\dfrac{21}{100})\\ \\ =2.303\times 10^{6}m^{-1}\)

Wavelength (λ) of the emitted radiation is:

\(\lambda =\dfrac{1}{\nu }\\ \\ =\dfrac{1}{2.303\times 10^{6}m^{-1}}\\ \\ =0.434\times 10^{-6}m\, \lambda \\ \\ =434nm\)

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?