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A photon collides with a stationary hydrogen atom in ground state inelastically. Energy of the colliding photon is 10.2 eV. After a time interval of the order of micro second another photon collides with same hydrogen atom inelastically with an energy of 15 eV. What will be observed by the detector ?
- a)2 photon of energy 10.2 eV
- b)2 photon of energy 1.4 eV
- c)one photon of energy 10.2 eV and an electron of energy 1.4 eV
- d)one photon of energy 10.2 eV and another photon of 1.4 eV
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
A photon collides with a stationary hydrogen atom in ground state ine...
Explanation:
Inelastic collision refers to a collision in which there is a transfer of energy between the colliding particles. In this case, a photon collides with a hydrogen atom in ground state inelastically, and then after a time interval of the order of microsecond, another photon collides with the same hydrogen atom inelastically. We need to determine what will be observed by the detector.
Step 1: Photon collides with hydrogen atom inelastically with an energy of 10.2 eV
When a photon collides with a hydrogen atom inelastically, the photon's energy is transferred to the hydrogen atom, exciting it to a higher energy level. In this case, the energy of the colliding photon is 10.2 eV, which is greater than the ionization energy of hydrogen (13.6 eV) but less than the energy required to excite hydrogen to the next energy level (12.1 eV). Therefore, the hydrogen atom will be excited to a higher energy level, but not to the next energy level.
Step 2: Another photon collides with the same hydrogen atom inelastically with an energy of 15 eV
After a time interval of the order of microsecond, another photon collides with the same hydrogen atom inelastically. The energy of the colliding photon is 15 eV, which is greater than the energy required to excite hydrogen to the next energy level (12.1 eV). Therefore, the hydrogen atom will be excited to the next energy level.
Step 3: What will be observed by the detector?
When the hydrogen atom is excited to a higher energy level by the first photon, it will eventually emit a photon to return to its ground state. The energy of this emitted photon will be equal to the energy difference between the excited state and the ground state. In this case, the energy difference is 10.2 eV, so one photon of energy 10.2 eV will be observed by the detector.
When the hydrogen atom is excited to the next energy level by the second photon, it will eventually emit a photon to return to its ground state. The energy of this emitted photon will be equal to the energy difference between the excited state and the ground state. In this case, the energy difference is 1.4 eV, so one photon of energy 1.4 eV will be observed by the detector.
In addition to the emitted photons, the collision of the second photon with the hydrogen atom will also result in the ejection of an electron. The energy of this ejected electron will be equal to the difference between the energy of the colliding photon (15 eV) and the ionization energy of hydrogen (13.6 eV), which is 1.4 eV. Therefore, an electron of energy 1.4 eV will also be observed by the detector.
Therefore, the correct answer is option 'C': one photon of energy 10.2 eV and an electron of energy 1.4 eV will be observed by the detector.
Inelastic collision refers to a collision in which there is a transfer of energy between the colliding particles. In this case, a photon collides with a hydrogen atom in ground state inelastically, and then after a time interval of the order of microsecond, another photon collides with the same hydrogen atom inelastically. We need to determine what will be observed by the detector.
Step 1: Photon collides with hydrogen atom inelastically with an energy of 10.2 eV
When a photon collides with a hydrogen atom inelastically, the photon's energy is transferred to the hydrogen atom, exciting it to a higher energy level. In this case, the energy of the colliding photon is 10.2 eV, which is greater than the ionization energy of hydrogen (13.6 eV) but less than the energy required to excite hydrogen to the next energy level (12.1 eV). Therefore, the hydrogen atom will be excited to a higher energy level, but not to the next energy level.
Step 2: Another photon collides with the same hydrogen atom inelastically with an energy of 15 eV
After a time interval of the order of microsecond, another photon collides with the same hydrogen atom inelastically. The energy of the colliding photon is 15 eV, which is greater than the energy required to excite hydrogen to the next energy level (12.1 eV). Therefore, the hydrogen atom will be excited to the next energy level.
Step 3: What will be observed by the detector?
When the hydrogen atom is excited to a higher energy level by the first photon, it will eventually emit a photon to return to its ground state. The energy of this emitted photon will be equal to the energy difference between the excited state and the ground state. In this case, the energy difference is 10.2 eV, so one photon of energy 10.2 eV will be observed by the detector.
When the hydrogen atom is excited to the next energy level by the second photon, it will eventually emit a photon to return to its ground state. The energy of this emitted photon will be equal to the energy difference between the excited state and the ground state. In this case, the energy difference is 1.4 eV, so one photon of energy 1.4 eV will be observed by the detector.
In addition to the emitted photons, the collision of the second photon with the hydrogen atom will also result in the ejection of an electron. The energy of this ejected electron will be equal to the difference between the energy of the colliding photon (15 eV) and the ionization energy of hydrogen (13.6 eV), which is 1.4 eV. Therefore, an electron of energy 1.4 eV will also be observed by the detector.
Therefore, the correct answer is option 'C': one photon of energy 10.2 eV and an electron of energy 1.4 eV will be observed by the detector.
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Community Answer
A photon collides with a stationary hydrogen atom in ground state ine...
13.6 eV energy needed to liberate the electron form hydrogen atom. So electron will liberate with kinetic energy. = 15 - 13.6 = 1.4 eV.
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A photon collides with a stationary hydrogen atom in ground state inelastically. Energy of the colliding photon is 10.2 eV. After a time interval of the order of micro second another photon collides with same hydrogen atom inelastically with an energy of 15 eV. What will be observed by the detector ?a)2 photon of energy 10.2 eVb)2 photon of energy 1.4 eVc)one photon of energy 10.2 eV and an electron of energy 1.4 eVd)one photon of energy 10.2 eV and another photon of 1.4 eVCorrect answer is option 'C'. Can you explain this answer?
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A photon collides with a stationary hydrogen atom in ground state inelastically. Energy of the colliding photon is 10.2 eV. After a time interval of the order of micro second another photon collides with same hydrogen atom inelastically with an energy of 15 eV. What will be observed by the detector ?a)2 photon of energy 10.2 eVb)2 photon of energy 1.4 eVc)one photon of energy 10.2 eV and an electron of energy 1.4 eVd)one photon of energy 10.2 eV and another photon of 1.4 eVCorrect answer is option 'C'. 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 photon collides with a stationary hydrogen atom in ground state inelastically. Energy of the colliding photon is 10.2 eV. After a time interval of the order of micro second another photon collides with same hydrogen atom inelastically with an energy of 15 eV. What will be observed by the detector ?a)2 photon of energy 10.2 eVb)2 photon of energy 1.4 eVc)one photon of energy 10.2 eV and an electron of energy 1.4 eVd)one photon of energy 10.2 eV and another photon of 1.4 eVCorrect answer is option 'C'. 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 photon collides with a stationary hydrogen atom in ground state inelastically. Energy of the colliding photon is 10.2 eV. After a time interval of the order of micro second another photon collides with same hydrogen atom inelastically with an energy of 15 eV. What will be observed by the detector ?a)2 photon of energy 10.2 eVb)2 photon of energy 1.4 eVc)one photon of energy 10.2 eV and an electron of energy 1.4 eVd)one photon of energy 10.2 eV and another photon of 1.4 eVCorrect answer is option 'C'. Can you explain this answer?.
A photon collides with a stationary hydrogen atom in ground state inelastically. Energy of the colliding photon is 10.2 eV. After a time interval of the order of micro second another photon collides with same hydrogen atom inelastically with an energy of 15 eV. What will be observed by the detector ?a)2 photon of energy 10.2 eVb)2 photon of energy 1.4 eVc)one photon of energy 10.2 eV and an electron of energy 1.4 eVd)one photon of energy 10.2 eV and another photon of 1.4 eVCorrect answer is option 'C'. 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 photon collides with a stationary hydrogen atom in ground state inelastically. Energy of the colliding photon is 10.2 eV. After a time interval of the order of micro second another photon collides with same hydrogen atom inelastically with an energy of 15 eV. What will be observed by the detector ?a)2 photon of energy 10.2 eVb)2 photon of energy 1.4 eVc)one photon of energy 10.2 eV and an electron of energy 1.4 eVd)one photon of energy 10.2 eV and another photon of 1.4 eVCorrect answer is option 'C'. 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 photon collides with a stationary hydrogen atom in ground state inelastically. Energy of the colliding photon is 10.2 eV. After a time interval of the order of micro second another photon collides with same hydrogen atom inelastically with an energy of 15 eV. What will be observed by the detector ?a)2 photon of energy 10.2 eVb)2 photon of energy 1.4 eVc)one photon of energy 10.2 eV and an electron of energy 1.4 eVd)one photon of energy 10.2 eV and another photon of 1.4 eVCorrect answer is option 'C'. Can you explain this answer?.
Solutions for A photon collides with a stationary hydrogen atom in ground state inelastically. Energy of the colliding photon is 10.2 eV. After a time interval of the order of micro second another photon collides with same hydrogen atom inelastically with an energy of 15 eV. What will be observed by the detector ?a)2 photon of energy 10.2 eVb)2 photon of energy 1.4 eVc)one photon of energy 10.2 eV and an electron of energy 1.4 eVd)one photon of energy 10.2 eV and another photon of 1.4 eVCorrect answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for NEET.
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Here you can find the meaning of A photon collides with a stationary hydrogen atom in ground state inelastically. Energy of the colliding photon is 10.2 eV. After a time interval of the order of micro second another photon collides with same hydrogen atom inelastically with an energy of 15 eV. What will be observed by the detector ?a)2 photon of energy 10.2 eVb)2 photon of energy 1.4 eVc)one photon of energy 10.2 eV and an electron of energy 1.4 eVd)one photon of energy 10.2 eV and another photon of 1.4 eVCorrect answer is option 'C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
A photon collides with a stationary hydrogen atom in ground state inelastically. Energy of the colliding photon is 10.2 eV. After a time interval of the order of micro second another photon collides with same hydrogen atom inelastically with an energy of 15 eV. What will be observed by the detector ?a)2 photon of energy 10.2 eVb)2 photon of energy 1.4 eVc)one photon of energy 10.2 eV and an electron of energy 1.4 eVd)one photon of energy 10.2 eV and another photon of 1.4 eVCorrect answer is option 'C'. Can you explain this answer?, a detailed solution for A photon collides with a stationary hydrogen atom in ground state inelastically. Energy of the colliding photon is 10.2 eV. After a time interval of the order of micro second another photon collides with same hydrogen atom inelastically with an energy of 15 eV. What will be observed by the detector ?a)2 photon of energy 10.2 eVb)2 photon of energy 1.4 eVc)one photon of energy 10.2 eV and an electron of energy 1.4 eVd)one photon of energy 10.2 eV and another photon of 1.4 eVCorrect answer is option 'C'. Can you explain this answer? has been provided alongside types of A photon collides with a stationary hydrogen atom in ground state inelastically. Energy of the colliding photon is 10.2 eV. After a time interval of the order of micro second another photon collides with same hydrogen atom inelastically with an energy of 15 eV. What will be observed by the detector ?a)2 photon of energy 10.2 eVb)2 photon of energy 1.4 eVc)one photon of energy 10.2 eV and an electron of energy 1.4 eVd)one photon of energy 10.2 eV and another photon of 1.4 eVCorrect answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A photon collides with a stationary hydrogen atom in ground state inelastically. Energy of the colliding photon is 10.2 eV. After a time interval of the order of micro second another photon collides with same hydrogen atom inelastically with an energy of 15 eV. What will be observed by the detector ?a)2 photon of energy 10.2 eVb)2 photon of energy 1.4 eVc)one photon of energy 10.2 eV and an electron of energy 1.4 eVd)one photon of energy 10.2 eV and another photon of 1.4 eVCorrect answer is option 'C'. Can you explain this answer? tests, examples and also practice NEET tests.
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