esiLimelawcuu

2023-02-25

Calculate the wavelength in Angstrom of the photon emitted when an electron in the Bohr orbit, n = 2 returns to the orbit, n = 1 in the hydrogen atom. The ionization energy of the ground state hydrogen atom is $2.17×{10}^{-11}$ erg per atom. (Take $h=6.62×{10}^{-27}erg-s$)

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Given,
Returns from the Bohr orbit of electrons state.
The hydrogen atom's ground state ionization energy $=2.17×{10}^{-11}erg/atom$
Again, We know
Ionisation energy $=-{E}_{1}$
i.e ionisation energy is the energy required to remove an electron from ground state, so indirectly we are provided with the value of ${E}_{1}$.
Also, ${E}_{2}-{E}_{1}=\Delta E$(when an electron returns from 2 to 1)$=\frac{hc}{\lambda }$
$⇒\lambda =\frac{hc}{{E}_{2}-{E}_{1}}$ ...(1)
Where, h = Planck's constant
c = speed of light
We know,
${E}_{n}\propto \frac{{Z}^{2}}{{n}^{2}}$
[Here Z is constant, since we are considering the same atom]
Hence,
${E}_{n}\propto \frac{1}{{n}^{2}}$
$\frac{{E}_{1}}{{E}_{n}}=\frac{{n}^{2}}{1}$
We can say that,
${E}_{n}=\frac{{E}_{1}}{{n}^{2}}$
Hence putting n = 2,
$⇒{E}_{2}=\frac{{E}_{1}}{4}=\frac{-2.17×{10}^{-11}}{4}$ erg/atom$=-0.5425×{10}^{-11}$ erg/atom
So,
$\Delta E={E}_{2}-{E}_{1}=-0.5425×{10}^{-11}erg-\left(-2.17×{10}^{-11}\right)=1.6275×{10}^{-11}erg$
∴ from equation (1),
$\lambda =\frac{\left(6.62×{10}^{-27}\right)×\left(3×{10}^{10}\right)}{1.6275×{10}^{-11}}=12.20×{10}^{-6}cm={1220}^{o}A$

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