Q2 An absorption line going from n1 to n2 has the same energy as an emission line from n2 to n1. So series in absorption correspond to emission series, but are defined by a common initial state rather than final state. So which series do you see in absorption?

Q3 How do you get two values for the wavefunction? Inserting a value of r should only give you one. It's not exactly 1.90; solve the equation for the nodes.

Q4 Still makes no sense, except that the ionisation energy of H is 13.6 eV. But atoms don't normally absorb or emit more than 1 electron at a time. Can you quote the whole question, with the multiple choice answers, exactly please.

Q5 The approach looks correct; please show your calculation so we can see if you made a mistake or (like Q1) the book got it wrong.

Q2. oh YEAH! In emission spectrum, the emission lines are recorded. Lyman series is bound to come. In absorption spectrum, the dark lines are the lines of wavelengths which are absorbed. So, Lyman series cannot be dark. So, Lyman series is obtained in both Absorption Spectra and emission Spectra. Thank you! Thanks a lot, mjc123.

Q.3.You mean to say, it is not 1.99 a

_{o} , rather 1.9 alone? When I said I got 2 values for the node, I substituted x = 1.99 a

_{o} , (and the a

_{o} terms in the numerator and the denominator got cancelled) and I was left with a quadratic in r. And both values of r were irrational and positive.

**Q4. A hydrogen like atom in ground state absorbs n photons having the same energy and it emits exactly n photons when electronic transition takes place. Then the energy of the absorbed photon is given by? (Multiple Correct)**

a) 91.8 eV b) 40.8 eV c) 48.4 eV d) 54.4 eVQ5. Fine. Here it goes.

1keV = 1.6 x 10

^{-16} J = K.E

So, [tex] p = \sqrt{2mK.E} [/tex]

Where, m = 9.1 x 10

^{-31} kg.

So,

[tex] p = \sqrt{2*9.1* 10^{-31} * 1.6 * 10^{-16}} [/tex]

[tex] p = \sqrt{29.12*10^{-47}} [/tex]

[tex] p = \sqrt{2.912*10^{-46}} [/tex]

[tex] p = 1.706*10^{-23} m/s [/tex]

Using Heisenberg's uncertainty principle, and assuming the minimum case:

[tex] \Delta x \Delta p = \frac{h}{4\pi} [/tex] So,

[tex] \Delta p = \frac{h}{4\pi\Delta x} [/tex]

[tex] \Delta p = \frac{6.62*10^{-34}}{4*3.14*10^{-10}} [/tex]

It is given that the position is located to within 1

^{o}. After simplifications, we get,

[tex] \Delta p = 0.527*10^{-24} m/s [/tex]

My assumption,

[tex] \Delta p = \frac{x}{100}*p [/tex]

[tex] x = 100\frac{\Delta p}{p} [/tex]

[tex] x = 100\frac{0.527*10^{-24}}{17.06*10^{-24}} [/tex]

[tex] x = \frac{52.7}{17.06}[/tex]

[tex] x = 3.08 [/tex]

But the answer given in the text was 6.178. Kindly point out the mistakes (if any) in my approach.Judging by my Q.1, this book isn't much of reliable!

Thank you!