Atomic Structure - Result Question 106
####82. Estimate the difference in energy between 1 st and 2 nd Bohr’s orbit for a hydrogen atom. At what minimum atomic number, a transition from $n=2$ to $n=1$ energy level would result in the emission of X-rays with $l=3.0 \times 10^{-8} m$ ? Which hydrogen atom-like species does this atomic number correspond to?
$(1993,5 M)$
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Solution:
- For H-atom, the energy of a stationary orbit is determined as
$$ \begin{aligned} & E _n=-\frac{k}{n^{2}} \quad \text { where, } k=\text { constant }\left(2.18 \times 10^{-18} J\right) \ \Rightarrow & \Delta E(n=2 \text { to } n=1)=k\left(1-\frac{1}{4}\right)=\frac{3}{4} k \ = & 1.635 \times 10^{-18} J \end{aligned} $$
For a H-like species, energy of stationary orbit is determined as
$$ E _n=-\frac{k Z^{2}}{n^{2}} $$
where, $Z=$ atomic number
$$ \begin{aligned} \Rightarrow & & \Delta E & =k Z^{2}\left(\frac{1}{n _1^{2}}-\frac{1}{n _2^{2}}\right) \ \Rightarrow & & \frac{1}{\lambda} & =\frac{\Delta E}{h c}=\frac{k}{h c} Z^{2}\left(\frac{1}{1}-\frac{1}{4}\right)=R _H Z^{2} \times \frac{3}{4} \ \Rightarrow & & Z^{2} & =\frac{4}{3 R _H \lambda}=\frac{4}{3 \times 1.097 \times 10^{7} \times 3 \times 10^{-8}}=4.05 \ \Rightarrow & & Z & =2\left(He^{+}\right) \end{aligned} $$
