Question

A particular hydrogen like atom has its ground state binding "energy 122.4 eV. Its is in ground state. Then

• A. Its atomic number is 3
• B. An electron of 90eV can excite it.
• C. An electron of kinetic energy nearly 91.8 eV can be brought to almost rest by this atom.
• D. An electron of kinetic energy 2.6 eV may emerge from the atom when electron of kinetic energy 125 eV collides with this atom.

Answer 1

Answer: (A), (C), (D)

A, C, D

Answer 2

1. Determine the atomic number (Z):

The ground state binding energy of a hydrogen-like atom is given by $|E_1| = 13.6 \frac{Z^2}{n^2}$ eV. For the ground state, n=1. Given binding energy = 122.4 eV. $13.6 Z^2 = 122.4$ $Z^2 = \frac{122.4}{13.6} = 9$ $Z = 3$ (since atomic number must be positive). So, the atomic number of the atom is 3.

2. Calculate the energy levels:

The energy levels for this atom are given by $E_n = -13.6 \frac{Z^2}{n^2} = -13.6 \frac{3^2}{n^2} = -\frac{122.4}{n^2}$ eV.

• Ground state (n=1): $E_1 = -122.4$ eV
• First excited state (n=2): $E_2 = -\frac{122.4}{2^2} = -\frac{122.4}{4} = -30.6$ eV
• Second excited state (n=3): $E_3 = -\frac{122.4}{3^2} = -\frac{122.4}{9} = -13.6$ eV
• Ionization energy (n=$\infty$): $E_\infty = 0$ eV

Now, let's analyze each option:

(A) Its atomic number is 3. As calculated above, Z=3. Therefore, option (A) is correct.

(B) An electron of 90 eV can excite it. The minimum energy required to excite the atom from its ground state (n=1) to the first excited state (n=2) is: $\Delta E_{12} = E_2 - E_1 = -30.6 \text{ eV} - (-122.4 \text{ eV}) = 91.8 \text{ eV}$. Since the incoming electron has 90 eV kinetic energy, which is less than 91.8 eV, it cannot excite the atom. Therefore, option (B) is incorrect.

(C) An electron of kinetic energy nearly 91.8 eV can be brought to almost rest by this atom. If an electron with kinetic energy 91.8 eV collides with the atom, it can transfer all its energy to excite the atom from the ground state (n=1) to the first excited state (n=2). In this process, the incident electron can lose all its kinetic energy and be brought to rest (assuming negligible recoil of the atom). "Nearly 91.8 eV" implies that it is very close to the exact excitation energy. Therefore, option (C) is correct.

(D) An electron of kinetic energy 2.6 eV may emerge from the atom when an electron of kinetic energy 125 eV collides with this atom. When an electron of kinetic energy 125 eV collides with the atom in its ground state: The energy required to ionize the atom from its ground state is its binding energy, which is 122.4 eV. If the atom is ionized, the incident electron transfers 122.4 eV to remove the electron. The remaining kinetic energy is $125 \text{ eV} - 122.4 \text{ eV} = 2.6 \text{ eV}$. This remaining 2.6 eV kinetic energy can be shared between the scattered incident electron and the newly ejected electron. It is possible that the incident electron emerges with 2.6 eV kinetic energy, or the ejected electron emerges with 2.6 eV kinetic energy, or the energy is shared between them. Since the question states "may emerge", it refers to one of these possibilities. Therefore, option (D) is correct.