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A particular hydrogen like atom has its ground state binding energy = 122.4 eV. It is in ground state. Then,
- a)its atomic number is 5
- b)an electron of 90 eV can excite it
- c)an electron of kinetic energy 45.9 eV can be brought to almost rest by this atom
- d)an electron of kinetic energy nearly 2.6 eV may emerge from the atom when electron of kinetic energy 125 eV collides with this atom
Correct answer is option 'D'. Can you explain this answer?
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A particular hydrogen like atom has its ground state binding energy =...
Given information:
- Ground state binding energy of the hydrogen-like atom = 122.4 eV
To determine the atomic number, we need to use the formula for the binding energy of a hydrogen-like atom:
Binding Energy = 13.6 * (Z^2 / n^2) eV
Where Z is the atomic number and n is the principal quantum number.
a) Determining the atomic number:
Using the formula, we can rearrange it to solve for Z:
Z^2 = (Binding Energy * n^2) / 13.6
For the ground state, n = 1. Plugging in the values:
Z^2 = (122.4 * 1^2) / 13.6
Z^2 = 9
Z = 3
Therefore, the atomic number of the hydrogen-like atom is 3.
b) Excitation of the atom by an electron with 90 eV kinetic energy:
To excite the atom, the electron must gain energy equal to the difference between the higher energy level and the ground state.
Since we know the binding energy of the ground state is 122.4 eV, the electron must gain an additional energy of 90 eV to be excited.
Therefore, an electron with 90 eV kinetic energy can excite the atom.
c) Bringing an electron with 45.9 eV kinetic energy to rest:
To bring an electron to rest, it must lose all of its kinetic energy. This can occur through collision with another particle, such as the hydrogen-like atom.
Since the kinetic energy of the electron is 45.9 eV, which is less than the binding energy of the ground state (122.4 eV), the electron can transfer its kinetic energy to the atom and come to rest.
Therefore, an electron with 45.9 eV kinetic energy can be brought to rest by this atom.
d) Emission of an electron with 2.6 eV kinetic energy:
When an electron with 125 eV kinetic energy collides with the atom, it can transfer some of its energy to the atom. If the transferred energy is less than the binding energy of the ground state, the atom remains in its ground state.
Since 125 eV - 122.4 eV = 2.6 eV, some of the kinetic energy can be transferred to the atom, and an electron with approximately 2.6 eV kinetic energy may emerge from the atom.
Therefore, an electron with kinetic energy nearly 2.6 eV may emerge from the atom when an electron with kinetic energy 125 eV collides with this atom.
- Ground state binding energy of the hydrogen-like atom = 122.4 eV
To determine the atomic number, we need to use the formula for the binding energy of a hydrogen-like atom:
Binding Energy = 13.6 * (Z^2 / n^2) eV
Where Z is the atomic number and n is the principal quantum number.
a) Determining the atomic number:
Using the formula, we can rearrange it to solve for Z:
Z^2 = (Binding Energy * n^2) / 13.6
For the ground state, n = 1. Plugging in the values:
Z^2 = (122.4 * 1^2) / 13.6
Z^2 = 9
Z = 3
Therefore, the atomic number of the hydrogen-like atom is 3.
b) Excitation of the atom by an electron with 90 eV kinetic energy:
To excite the atom, the electron must gain energy equal to the difference between the higher energy level and the ground state.
Since we know the binding energy of the ground state is 122.4 eV, the electron must gain an additional energy of 90 eV to be excited.
Therefore, an electron with 90 eV kinetic energy can excite the atom.
c) Bringing an electron with 45.9 eV kinetic energy to rest:
To bring an electron to rest, it must lose all of its kinetic energy. This can occur through collision with another particle, such as the hydrogen-like atom.
Since the kinetic energy of the electron is 45.9 eV, which is less than the binding energy of the ground state (122.4 eV), the electron can transfer its kinetic energy to the atom and come to rest.
Therefore, an electron with 45.9 eV kinetic energy can be brought to rest by this atom.
d) Emission of an electron with 2.6 eV kinetic energy:
When an electron with 125 eV kinetic energy collides with the atom, it can transfer some of its energy to the atom. If the transferred energy is less than the binding energy of the ground state, the atom remains in its ground state.
Since 125 eV - 122.4 eV = 2.6 eV, some of the kinetic energy can be transferred to the atom, and an electron with approximately 2.6 eV kinetic energy may emerge from the atom.
Therefore, an electron with kinetic energy nearly 2.6 eV may emerge from the atom when an electron with kinetic energy 125 eV collides with this atom.
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Community Answer
A particular hydrogen like atom has its ground state binding energy =...
Ground state binding energy = 13.6 × z2 eV
The first excitation energy
K.Emax = 125 − 122.4 = 2.6 eV
Hence, 90 eV90 eV electron cannot excite it.
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A particular hydrogen like atom has its ground state binding energy = 122.4 eV. It is in ground state. Then,a)its atomic number is 5b)an electron of 90 eV can excite itc)an electron of kinetic energy 45.9 eV can be brought to almost rest by this atomd)an electron of kinetic energy nearly 2.6 eV may emerge from the atom when electron of kinetic energy 125 eV collides with this atomCorrect answer is option 'D'. Can you explain this answer?
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A particular hydrogen like atom has its ground state binding energy = 122.4 eV. It is in ground state. Then,a)its atomic number is 5b)an electron of 90 eV can excite itc)an electron of kinetic energy 45.9 eV can be brought to almost rest by this atomd)an electron of kinetic energy nearly 2.6 eV may emerge from the atom when electron of kinetic energy 125 eV collides with this atomCorrect answer is option 'D'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A particular hydrogen like atom has its ground state binding energy = 122.4 eV. It is in ground state. Then,a)its atomic number is 5b)an electron of 90 eV can excite itc)an electron of kinetic energy 45.9 eV can be brought to almost rest by this atomd)an electron of kinetic energy nearly 2.6 eV may emerge from the atom when electron of kinetic energy 125 eV collides with this atomCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particular hydrogen like atom has its ground state binding energy = 122.4 eV. It is in ground state. Then,a)its atomic number is 5b)an electron of 90 eV can excite itc)an electron of kinetic energy 45.9 eV can be brought to almost rest by this atomd)an electron of kinetic energy nearly 2.6 eV may emerge from the atom when electron of kinetic energy 125 eV collides with this atomCorrect answer is option 'D'. Can you explain this answer?.
A particular hydrogen like atom has its ground state binding energy = 122.4 eV. It is in ground state. Then,a)its atomic number is 5b)an electron of 90 eV can excite itc)an electron of kinetic energy 45.9 eV can be brought to almost rest by this atomd)an electron of kinetic energy nearly 2.6 eV may emerge from the atom when electron of kinetic energy 125 eV collides with this atomCorrect answer is option 'D'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A particular hydrogen like atom has its ground state binding energy = 122.4 eV. It is in ground state. Then,a)its atomic number is 5b)an electron of 90 eV can excite itc)an electron of kinetic energy 45.9 eV can be brought to almost rest by this atomd)an electron of kinetic energy nearly 2.6 eV may emerge from the atom when electron of kinetic energy 125 eV collides with this atomCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particular hydrogen like atom has its ground state binding energy = 122.4 eV. It is in ground state. Then,a)its atomic number is 5b)an electron of 90 eV can excite itc)an electron of kinetic energy 45.9 eV can be brought to almost rest by this atomd)an electron of kinetic energy nearly 2.6 eV may emerge from the atom when electron of kinetic energy 125 eV collides with this atomCorrect answer is option 'D'. Can you explain this answer?.
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A particular hydrogen like atom has its ground state binding energy = 122.4 eV. It is in ground state. Then,a)its atomic number is 5b)an electron of 90 eV can excite itc)an electron of kinetic energy 45.9 eV can be brought to almost rest by this atomd)an electron of kinetic energy nearly 2.6 eV may emerge from the atom when electron of kinetic energy 125 eV collides with this atomCorrect answer is option 'D'. Can you explain this answer?, a detailed solution for A particular hydrogen like atom has its ground state binding energy = 122.4 eV. It is in ground state. Then,a)its atomic number is 5b)an electron of 90 eV can excite itc)an electron of kinetic energy 45.9 eV can be brought to almost rest by this atomd)an electron of kinetic energy nearly 2.6 eV may emerge from the atom when electron of kinetic energy 125 eV collides with this atomCorrect answer is option 'D'. Can you explain this answer? has been provided alongside types of A particular hydrogen like atom has its ground state binding energy = 122.4 eV. It is in ground state. Then,a)its atomic number is 5b)an electron of 90 eV can excite itc)an electron of kinetic energy 45.9 eV can be brought to almost rest by this atomd)an electron of kinetic energy nearly 2.6 eV may emerge from the atom when electron of kinetic energy 125 eV collides with this atomCorrect answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
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