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conantee
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« on: December 31, 2009, 01:04:31 PM »

26. A nickel target (Z = 28) is bombarded with fast electrons. The minimum electron kinetic energy needed to produce x-rays in the K serires is most nearly
(A) 10 \mbox{ eV}
(B) 100 \mbox{ eV}
(C) 1000 \mbox{ eV}
(D) 10,000 \mbox{ eV}
(E) 100,000 \mbox{ eV}
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« Reply #1 on: December 31, 2009, 01:29:48 PM »

This requires more knowledge about atomic structure in multielectron atoms. Still, we duplicate the logic from hydrogen atom.

Now, let's get back to hydrogen atom, we know that energy of hydrogen atom is
E_n = - \frac{13.6 \mbox { eV}}{n^2} for n =1,2,3,...

Now, for inner most level, the electron in nickel's atom "sees" an effective nuclear charge of approximately (Z-1)e, where e is elementary charge and Z is the atomic number of the element. This is because there are two electrons in the deepest shell. One will see nucleus and another electron be an effective nucleus. If you remember Bohr's theory and how to derive energy levels, you should notice that the charge of nucleus appears as square in the energy term. So, the energy is revised to
E_n = - \frac{(13.6 \mbox { eV})(Z-1)^2}{n^2} for n =1,2,3,...

Now, in order to produce x-rays in the K series, i.e., x-rays from transition of n=1 level, the minimum energy is from transition from n=1 to the closest level, n =2.
E_{min} = \frac{(13.6 \mbox { eV})(Z-1)^2}{1^2} - \frac{(13.6 \mbox { eV})(Z-1)^2}{2^2} = (13.6 \mbox{ eV})(Z-1)^2\frac{3}{4}

Plugging Z =28 , we got
E_{min}  = (13.6 \mbox{ eV})(27)^2\frac{3}{4} = 3.4 \cdot (27)^2 \cdot 3 \approx 3 \cdot 3^6 \cdot 3 \approx 3^8 \approx (10^{0.5})^8 = 10^4 \mbox{ eV}

The correct is therefore (D). 28 of 100 people got the right answer.
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