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 $2.17\times {10}^{-11}$ erg per atom. (Take $h=6.62\times {10}^{-27}erg-s$)

tammypierce5kgk

Beginner2023-02-26Added 7 answers

Given,

Returns from the Bohr orbit of electrons ${n}_{2}=2\text{}to\text{}{n}_{1}=1$ state.

The hydrogen atom's ground state ionization energy $=2.17\times {10}^{-11}erg/atom$

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)$=\frac{hc}{\lambda}$

$\Rightarrow \lambda =\frac{hc}{{E}_{2}-{E}_{1}}$ ...(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,

$\Rightarrow {E}_{2}=\frac{{E}_{1}}{4}=\frac{-2.17\times {10}^{-11}}{4}$ erg/atom$=-0.5425\times {10}^{-11}$ erg/atom

So,

$\Delta E={E}_{2}-{E}_{1}=-0.5425\times {10}^{-11}erg-(-2.17\times {10}^{-11})=1.6275\times {10}^{-11}erg$

∴ from equation (1),

$\lambda =\frac{(6.62\times {10}^{-27})\times (3\times {10}^{10})}{1.6275\times {10}^{-11}}=12.20\times {10}^{-6}cm={1220}^{o}A$

Returns from the Bohr orbit of electrons ${n}_{2}=2\text{}to\text{}{n}_{1}=1$ state.

The hydrogen atom's ground state ionization energy $=2.17\times {10}^{-11}erg/atom$

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)$=\frac{hc}{\lambda}$

$\Rightarrow \lambda =\frac{hc}{{E}_{2}-{E}_{1}}$ ...(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,

$\Rightarrow {E}_{2}=\frac{{E}_{1}}{4}=\frac{-2.17\times {10}^{-11}}{4}$ erg/atom$=-0.5425\times {10}^{-11}$ erg/atom

So,

$\Delta E={E}_{2}-{E}_{1}=-0.5425\times {10}^{-11}erg-(-2.17\times {10}^{-11})=1.6275\times {10}^{-11}erg$

∴ from equation (1),

$\lambda =\frac{(6.62\times {10}^{-27})\times (3\times {10}^{10})}{1.6275\times {10}^{-11}}=12.20\times {10}^{-6}cm={1220}^{o}A$