According to Bohr theory- the electronic energy of hydrogen atom in the nth Bohr orbit is given by- E n - 21-76- 10 -19 n 2 JCalculate the longest wavelength of light that will be needed to remove an electron from the 2nd orbit of Li 2- ion-

E_n= 21.76× 10^ 19n^2J Energy at infinite distance =0 .Energy at infinite orbit = 21.76× 10^ 194J= 5.44× 10^ 19J Energy difference =5.44× 10^ ...

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Standard XII

Chemistry

Bohr's Model of a Hydrogen Atom

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Question

According to Bohr theory, the electronic energy of hydrogen atom in the $n$ th Bohr orbit is given by,
$E_{n} = - \frac{21.76 \times 10^{- 19}}{n^{2}} J$
Calculate the longest wavelength of light that will be needed to remove an electron from the $2$ nd orbit of ${L i}^{2 +}$ ion.

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Solution

$E_{n} = \frac{- 21.76 \times 10^{-} 19}{n^{2}} J$
Energy at infinite distance $= 0$ .
Energy at infinite orbit $= \frac{- 21.76 \times 10^{- 19}}{4} J = - 5.44 \times 10^{- 19} J$
Energy difference $= 5.44 \times 10^{- 19} = \frac{h c}{λ}$
$∴ λ = \frac{6.64 \times 10^{- 36} \times 3 \times 10^{8}}{5.44 \times 10^{- 19}} = 3.66 \times 10^{- 9} m$

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According to Bohr theory, the electronic energy of hydrogen atom in the nth Bohr orbit is given by, En=−21.76×10−19n2JCalculate the longest wavelength of light that will be needed to remove an electron from the 2nd orbit of Li2+ ion.

En=−21.76×10−19n2JEnergy at infinite distance =0.Energy at infinite orbit =−21.76×10−194J=−5.44×10−19JEnergy difference=5.44×10−19=hcλ∴λ=6.64×10−36×3×1085.44×10−19=3.66×10−9m

According to Bohr theory, the electronic energy of hydrogen atom in the $n$ th Bohr orbit is given by,
$E_{n} = - \frac{21.76 \times 10^{- 19}}{n^{2}} J$
Calculate the longest wavelength of light that will be needed to remove an electron from the $2$ nd orbit of ${L i}^{2 +}$ ion.