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(B) Let the ground state energy of the hydrogen-like atom be $E_0 = -13.6 Z^2 \ eV$. The energy of an electron in state $k$ is given by $E_k = \frac{E

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(B) Let the ground state energy of the hydrogen-like atom be $E_0 = -13.6 Z^2 \ eV$. The energy of an electron in state $k$ is given by $E_k = \frac{E_0}{k^2}$.Given that the atom is in state $2n$ and emits a maximum energy photon of $204 \ eV$ to reach the ground state $(k=1)$:$E_{2n} - E_1 = 204 \ eV$$\frac{E_0}{(2n)^2} - E_0 = 204 \ eV$$E_0 \left( \frac{1}{4n^2} - 1 \right) = 204 \ eV$  --- $(i)$Given that the atom makes a transition from state $2n$ to state $n$ emitting a photon of $40.8 \ eV$:$E_{2n} - E_n = 40.8 \ eV$$\frac{E_0}{4n^2} - \frac{E_0}{n^2} = 40.8 \ eV$$E_0 \left( \frac{1 - 4}{4n^2} \right) = 40.8 \ eV$$E_0 \left( -\frac{3}{4n^2} \right) = 40.8 \ eV$ --- $(ii)$Dividing equation $(i)$ by equation $(ii)$:$\frac{E_0 \left( \frac{1 - 4n^2}{4n^2} \right)}{E_0 \left( -\frac{3}{4n^2} \right)} = \frac{204}{40.8}$$\frac{1 - 4n^2}{-3} = 5$$1 - 4n^2 = -15$$4n^2 = 16$$n^2 = 4$$n = 2$

$A$ hydrogen-like atom of atomic number $Z$ is in an excited state of quantum number $2n$. It can emit a maximum energy photon of $204 \ eV$. If it makes a transition to quantum state $n$,a photon of energy $40.8 \ eV$ is emitted. The value of $n$ will be

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