Question
A single electron orbits around a stationary nucleus of charge +Ze in a hydrogen-like atom, where Z is the atomic number and e is the magnitude of the charge on an electron. It requires 47.25 eV to excite the electron from second Bohr orbit to the third Bohr orbit. Ionization energy of hydrogen atom is 13.6 eV. Then
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Solution
[Key Concept: Hydrogen-like energies are En = −13.6Z2/n2 eV, while Bohr angular momentum nh/(2π) does not depend on Z.]
a) $E_{3} - E_{2} = 47.25eV$
$$\begin{matrix} & \ \Rightarrow - 13.6Z^{2}\left( \frac{1}{9} - \frac{1}{4} \right) = 47.25 \\ & \ \Rightarrow Z^{2} = 25 \\ & \ \Rightarrow Z = 5 \end{matrix}$$
b) $E_{4} - E_{3} = - 13.6 \times 25 \times \left( \frac{1}{16} - \frac{1}{9} \right)$
$$= 16.53eV$$
c) $n = 1 \rightarrow n = \infty$
$$E = - 13.6 \times 25\left( \frac{1}{\infty} - \frac{1}{1} \right) = 340eV$$
Wavelength required $= \frac{12400}{340}\text{Å} = 36.56\text{Å}$
d) $L = \frac{nh}{2\pi} = \frac{2 \times 6.62 \times 10^{- 34}}{2 \times 3.14} = 2.19 \times 10^{- 34}$
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