Find the binding energy of valence electron in the ground state of a $$Li$$ atom if the wavelength of the sharp series is known to be $$\lambda_{1}$$ $$= 813 nm$$ and the short wave cutoff wavelength $$\lambda_{2}$$, $$= 350 nm$$
For the first line of sharp series $$(3s\rightarrow 2p)$$ in a $$Li$$ atom
$$\displaystyle \frac{hc}{\lambda}=\frac{hR}{(3+\alpha _0)^2} + \frac{hR}{(2+\alpha_1)^2} $$ ...........(1)
For the short wave cut off wavelength of the same series
$$\displaystyle \frac{hc}{\lambda_{2}}= \frac{hR}{(2+\alpha_1)^2} $$ ...............(2)
Subtracting (1) from (2), we get:
$$(3+\alpha _0)^2$$=\displaystyle \frac{hR\lambda_{1}\lambda_{2}}{hc(\lambda_{1}-\lambda_{2})}=\frac{R\lambda_{1}\lambda_{2}}{2\pi c(\lambda_{1}-\lambda_{2})}$$
$$\displaystyle 3+\alpha _{0}=\sqrt\frac{R\lambda _{1}\lambda _{2}}{2\pi c(\lambda _{1}-\lambda _{2})}$$
In the ground state BE of electrons is:
$$=\displaystyle \dfrac{hR}{(2+\alpha _{0})^{2}}$$
$$\displaystyle =\dfrac{hR}{\left (\sqrt{\dfrac{R\lambda_{1}\lambda_{2}}{2\pi c(\lambda_{1}-\lambda_{2})}-1} \right )^{2}}=5.32 eV$$
Assume that a neutron breaks into a proton and an electron. The energy released during this process is (Mass of neutron $$=1.6725\times 10^{-27} kg$$, Mass of proton $$=1.6725\times 10^{-27} kg$$ ,Mass of electron $$=9\times 10^{-31} kg$$)
The binding energy per nucleon for the parent nucleus is $$\mathrm{E}_{1}$$ and that for the daughter nuclei is $$\mathrm{E}_{2}$$.Then
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