Question

When energy is supplied to an atom, the electrons of the atom move to higher energy orbits depending on the amount of energy absorbed. When these electrons return to any of the lower energy orbits, they emit energy. Depending on whether the electrons return to the 1st, 2nd, 3rd, 4th or 5th orbits, the series formed are called Lyman, Balmer, Paschen, Brackett and Pfund series respectively.
If the electron comes back from the energy level $E_2$ to the energy level $E_1$, then the difference may be expressed in terms of the energy of photon as
$E_2-E_1=\Delta E$, $\lambda =\frac{hc}{\Delta E}$
Thus each transition from one energy level to another will produce a light of definite wavelength. This is observed as a line in the spectrum of the hydrogen atom.
Wave number of the line is given by the formula
$\stackrel{-}{\nu }=R{Z}^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})$
Where $R=1.097\times {10}^{5}c{m}^{-1}$ is Rydberg constant.
$\begin{array}{cc}\text{List - I}& \text{List - II}\\ \text{(I) If the ionisation potential for hydrogen-like atom}& \text{(P)}5.48\times {10}^{-32}cRJ\\ \text{in a sample is}122.0eV,\text{then the energy of the last}& \\ \text{line of Paschen series for this atom is:}& \\ \text{(II) In a single isolated atom, an electron makes}& \text{(Q)}6.6\times {10}^{-32}cRJ\\ \text{transition from fifth excited state such that no}& \\ \text{spectral lines are observed in the visible range}& \\ \text{then maximum number of different types of}& \\ \text{photons observed is:}& \\ \text{(III) The difference in energy of the second line}& \text{(R)}6\\ \text{of Lyman series and last line of Brackett series in a}& \\ \text{hydrogen sample is:}& \\ \text{(IV) The energy of the electromagnetic radiation}& \text{(S)}4\\ \text{emitted during the transition of electron in between}& \\ \text{the two levels of}L{i}^{2+}\text{ion whose principal quantum}& \\ \text{number sum is 4 and difference is 2 is:}& \\ & \text{(T)}6.6\times {10}^{-34}cRJ\\ & \text{(U)}0.53\times {10}^{-30}cRJ\end{array}$

Which of the following options has the correct combination considering List-I and List-II?

• A. $I\to Q,IV\to T$
• B. $II\to S,III\to P$
• C. $II\to R,III\to P$
• D. $I\to T,IV\to U$

Answer 1

The correct option is B $II\to S,III\to P$

a) Given, ionization potential for the hydrogen-like atom is $122.4eV$
So, $122.4=13.6\times Z^2$
$\Rightarrow Z=3$
So, $\stackrel{-}{\nu }=R{3}^2(\frac{1}{3^2}-0)=R$
$\therefore EnergyE=hc\stackrel{-}{\nu }=6.6\times {10}^{-34}cR\times {10}^{2}J=6.6\times {10}^{-32}cRJ$
b) The different possible transitions are $6\to 5,5\to 4,4\to 3,3\to 1,$
c) The wave number of the second line of Lyman series is
$\stackrel{-}{{\nu }_1}=R{1}^2(\frac{1}{1^2}-\frac{1}{3^2})=R\left(\frac{8}{9}\right)$
The wave number of the last line of Brackett series is
$\stackrel{-}{{\nu }_2}=R{1}^2(\frac{1}{4^2}-0)=R\left(\frac{1}{16}\right)$
So the energy difference will be $\Delta E=hc\Delta \nu =hc(\frac{8}{9}-\frac{1}{16})R\times {10}^2$
$=hcR\left(\frac{119}{144}\right)\times {10}^2=5.478\times {10}^{-32}cRJ$
d) From the question the electron is making transition from 3rd level to the 1st level. So the wave number of electromagnetic radiation will be
$\stackrel{-}{\nu }=R{3}^2(\frac{1}{1^2}-\frac{1}{3^2})=8R$
So the energy of the radiation will be
$E=6.6\times {10}^{-34}\times 8\times {10}^{2}cR=0.53\times {10}^{-30}cRJ$