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(B) The energy of an electron in the $n^{th}$ orbit is given by $E_n = -R_H 	imes \frac{Z^2}{n^2}$.The energy difference between two orbits $n_1$ and

Vedclass · Accepted Answer

$\frac{32}{27} 	imes \frac{3B}{4}$

Vedclass · Answer

(B) The energy of an electron in the $n^{th}$ orbit is given by $E_n = -R_H 	imes \frac{Z^2}{n^2}$.The energy difference between two orbits $n_1$ and $n_2$ is $\Delta E = R_H 	imes Z^2 	imes (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.For the first and second orbits $(n_1=1, n_2=2)$: $\Delta E_1 = R_H 	imes Z^2 	imes (1 - \frac{1}{4}) = R_H 	imes Z^2 	imes \frac{3}{4} = \frac{3B}{4}$.For the first and third orbits $(n_1=1, n_2=3)$: $\Delta E_2 = R_H 	imes Z^2 	imes (1 - \frac{1}{9}) = R_H 	imes Z^2 	imes \frac{8}{9}$.Taking the ratio: $\frac{\Delta E_2}{\Delta E_1} = \frac{8/9}{3/4} = \frac{8}{9} 	imes \frac{4}{3} = \frac{32}{27}$.Therefore,$\Delta E_2 = \frac{32}{27} 	imes \Delta E_1 = \frac{32}{27} 	imes \frac{3B}{4}$.

For a hydrogen atom,the energy difference between the first and second Bohr orbits is $3B/4$. What is the energy difference between the first and third Bohr orbits?

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