The radii of two of the first four Bohr’s orb | vedclass

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(B) The radius of the $n^{th}$ Bohr orbit is given by $r_n = a_0 n^2$. Given the ratio of radii is $\frac{r_{n_1}}{r_{n_2}} = \frac{n_1^2}{n_2^2} = \f

Vedclass · Accepted Answer

Either $2.55 \ eV$ or $10.2 \ eV$

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(B) The radius of the $n^{th}$ Bohr orbit is given by $r_n = a_0 n^2$. Given the ratio of radii is $\frac{r_{n_1}}{r_{n_2}} = \frac{n_1^2}{n_2^2} = \frac{1}{4}$,which implies $\frac{n_1}{n_2} = \frac{1}{2}$. Since the orbits are among the first four $(n = 1, 2, 3, 4)$,the possible pairs $(n_1, n_2)$ are $(1, 2)$ or $(2, 4)$. The energy of the $n^{th}$ orbit is $E_n = -\frac{13.6}{n^2} \ eV$. For the pair $(1, 2)$,the energy difference is $\Delta E = E_2 - E_1 = -\frac{13.6}{4} - (-13.6) = -3.4 + 13.6 = 10.2 \ eV$. For the pair $(2, 4)$,the energy difference is $\Delta E = E_4 - E_2 = -\frac{13.6}{16} - (-\frac{13.6}{4}) = -0.85 + 3.4 = 2.55 \ eV$.

The radii of two of the first four Bohr’s orbits of the hydrogen atom are in the ratio $1 : 4$. The energy difference between them may be

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