In a hydrogen-like atom with atomic number Z, | vedclass

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(B) The energy of an electron in the $m$-th orbit of a hydrogen-like atom is given by $E_m = \frac{E_1}{m^2}$,where $E_1$ is the ground state energy (

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$2$

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(B) The energy of an electron in the $m$-th orbit of a hydrogen-like atom is given by $E_m = \frac{E_1}{m^2}$,where $E_1$ is the ground state energy (which is negative).The maximum energy photon emitted from the $2n$-th state corresponds to a transition to the ground state $(m=1)$.Given: $E_{2n} - E_1 = 204 \ eV$.$\frac{E_1}{(2n)^2} - E_1 = 204 \ eV \implies E_1 \left( \frac{1}{4n^2} - 1 \right) = 204 \ eV \quad \dots(1)$When the electron transitions from the $2n$ orbit to the $n$ orbit,the energy of the emitted photon is $40.8 \ eV$.Given: $E_{2n} - E_n = 40.8 \ eV$.$\frac{E_1}{4n^2} - \frac{E_1}{n^2} = 40.8 \ eV \implies E_1 \left( \frac{1 - 4}{4n^2} \right) = 40.8 \ eV \implies E_1 \left( -\frac{3}{4n^2} \right) = 40.8 \ eV \quad \dots(2)$Dividing equation $(1)$ by equation $(2)$:$\frac{E_1 (\frac{1-4n^2}{4n^2})}{E_1 (-\frac{3}{4n^2})} = \frac{204}{40.8}$$\frac{1-4n^2}{-3} = 5 \implies 1 - 4n^2 = -15 \implies 4n^2 = 16 \implies n^2 = 4 \implies n = 2$.

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