\(\therefore \;\frac{{{{({W_0})}_T}}}{{{{({W_0})}_{Na}}}} = \frac{{{\lambda _{Na}}}}{{{\lambda _T}}}\) or \({\lambda _T} = \frac{{{\lambda _{Na}} \times {{({W_0})}_{Na}}}}{{{{({W_0})}_T}}}\)\( = \frac{{5460 \times 2.3}}{{4.5}} = 2791\;{Å}\)
$\left(\mathrm{m}_{\mathrm{e}}=9 \times 10^{-31}\;\mathrm{kg}\right)$