Question

Ground state energy of hydrogen atom is $-E_0$. In a single electron ion with atomic number $Z$, electron excited to $n^{th}$ quantum state makes transition to ground state by emitting two photons of energies $\frac{3E_0}{4}$ and $3E_0$. Also it may transit to first excited state by emitting two photons of energies $\frac{7E_0}{36}$ and $\frac{5E_0}{9}$. Then sum; $n + Z$ is ____

Answer 1

6

Answer 2

The energy of an electron in the $k^{th}$ quantum state of a single-electron ion with atomic number $Z$ is given by $E_k = -E_0 \frac{Z^2}{k^2}$.

Scenario 1: Transition from state $n$ to the ground state (state 1). The total energy emitted is $\Delta E_1 = \frac{3E_0}{4} + 3E_0 = \frac{15E_0}{4}$. This energy difference is $E_n - E_1 = -E_0 \frac{Z^2}{n^2} - (-E_0 \frac{Z^2}{1^2}) = E_0 Z^2 (1 - \frac{1}{n^2})$. So, $E_0 Z^2 (1 - \frac{1}{n^2}) = \frac{15E_0}{4}$, which simplifies to $Z^2 \left(\frac{n^2 - 1}{n^2}\right) = \frac{15}{4}$ (Equation 1).

Scenario 2: Transition from state $n$ to the first excited state (state 2). The total energy emitted is $\Delta E_2 = \frac{7E_0}{36} + \frac{5E_0}{9} = \frac{7E_0 + 20E_0}{36} = \frac{27E_0}{36} = \frac{3E_0}{4}$. This energy difference is $E_n - E_2 = -E_0 \frac{Z^2}{n^2} - (-E_0 \frac{Z^2}{2^2}) = E_0 Z^2 (\frac{1}{4} - \frac{1}{n^2})$. So, $E_0 Z^2 (\frac{1}{4} - \frac{1}{n^2}) = \frac{3E_0}{4}$, which simplifies to $Z^2 \left(\frac{n^2 - 4}{4n^2}\right) = \frac{3}{4}$ (Equation 2).

Dividing Equation 1 by Equation 2: $\frac{Z^2 \left(\frac{n^2 - 1}{n^2}\right)}{Z^2 \left(\frac{n^2 - 4}{4n^2}\right)} = \frac{15/4}{3/4}$ $\frac{4(n^2 - 1)}{n^2 - 4} = 5$ $4n^2 - 4 = 5n^2 - 20$ $n^2 = 16 \implies n = 4$

Substitute $n=4$ into Equation 2: $Z^2 \left(\frac{1}{4} - \frac{1}{16}\right) = \frac{3}{4}$ $Z^2 \left(\frac{3}{16}\right) = \frac{3}{4}$ $Z^2 = 4 \implies Z = 2$

The sum $n + Z = 4 + 2 = 6$.