Question

In a hydrogen like sample electron is in 2nd excited state, the Binding energy of 4th state of this sample is 13.6 eV, then

• A. A 25 eV photon can set free the electron from the second excited state of this sample.
• B. 3 different types of photon will be observed if electrons make transition up to ground state from the second excited state
• C. If 23 eV photon is used then K. E. of the-ejected electron is 1 eV.
• D. 2nd line of Balmer series of this sample has same energy value as 1st excitation energy of H-atoms.

Answer 1

Answer: (A), (B)

(A), (B)

Answer 2

The energy of an electron in the $n$-th state of a hydrogen-like atom with atomic number $Z$ is given by $E_n = -13.6 \frac{Z^2}{n^2}$ eV. The binding energy of the $n$-th state is $|E_n|$.

Given that the binding energy of the 4th state ($n=4$) is 13.6 eV: $|E_4| = 13.6$ eV $|-13.6 \frac{Z^2}{4^2}| = 13.6$ $13.6 \frac{Z^2}{16} = 13.6$ $\frac{Z^2}{16} = 1 \implies Z^2 = 16 \implies Z = 4$ (since Z must be positive). The sample is a hydrogen-like ion with $Z=4$.

The energy levels for this sample are $E_n = -13.6 \frac{4^2}{n^2} = -13.6 \frac{16}{n^2} = -217.6 \frac{1}{n^2}$ eV. The electron is initially in the 2nd excited state, which corresponds to $n=3$. $E_1 = -217.6 \frac{1}{1^2} = -217.6$ eV $E_2 = -217.6 \frac{1}{2^2} = -54.4$ eV $E_3 = -217.6 \frac{1}{3^2} = -24.18$ eV (approx.) $E_4 = -217.6 \frac{1}{4^2} = -13.6$ eV

Let's evaluate the options:

(A) A 25 eV photon can set free the electron from the second excited state ($n=3$). The binding energy of the $n=3$ state is $|E_3| = |-24.18 \text{ eV}| = 24.18$ eV. To set the electron free, the energy of the absorbed photon must be greater than or equal to the binding energy. Since 25 eV > 24.18 eV, a 25 eV photon can set the electron free. The kinetic energy of the ejected electron would be $25 \text{ eV} - 24.18 \text{ eV} = 0.82$ eV. Option (A) is correct.

(B) 3 different types of photon will be observed if electrons make transition up to ground state from the second excited state. The electron is in the $n=3$ state. Transitions to lower states are possible: $n=3 \to n=2$: Energy $\Delta E_{32} = E_3 - E_2 = -24.18 - (-54.4) = 30.22$ eV. $n=3 \to n=1$: Energy $\Delta E_{31} = E_3 - E_1 = -24.18 - (-217.6) = 193.42$ eV. $n=2 \to n=1$: Energy $\Delta E_{21} = E_2 - E_1 = -54.4 - (-217.6) = 163.2$ eV. These three transitions ($3 \to 2$, $3 \to 1$, and $2 \to 1$) result in three different photon energies. Option (B) is correct.

(C) If 23 eV photon is used then K. E. of the-ejected electron is 1 eV. The electron is in the $n=3$ state with binding energy 24.18 eV. A 23 eV photon has less energy than the binding energy. Therefore, it cannot eject the electron from this state. Option (C) is incorrect.

(D) 2nd line of Balmer series of this sample has same energy value as 1st excitation energy of H-atoms. Balmer series transitions end at $n=2$. The 2nd line of Balmer series is the transition from $n=4$ to $n=2$. Energy of 2nd line of Balmer series for the sample ($Z=4$): $\Delta E_{4 \to 2} = E_4 - E_2 = (-13.6) - (-54.4) = 40.8$ eV.

1st excitation energy of H-atom ($Z=1$) is the energy required for the transition from $n=1$ to $n=2$. For H-atom, $E_n = -13.6 \frac{1^2}{n^2} = -13.6 \frac{1}{n^2}$ eV. $E_1 = -13.6$ eV, $E_2 = -13.6 \frac{1}{4} = -3.4$ eV. 1st excitation energy $\Delta E_{1 \to 2} = E_2 - E_1 = -3.4 - (-13.6) = 10.2$ eV. Comparing the energies: 40.8 eV $\ne$ 10.2 eV. Option (D) is incorrect.

Based on the analysis, options (A) and (B) are correct.