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An electron is in an excited state in hydrogen-like atom. It has a total energy of –3.4 eV. If the kinetic energy of the electron is E and its de-Broglie wavelength is l, then
- a)E = 6.8 eV, l = 6.6 × 10–10 m
- b)E = 3.4 eV, l = 6.6 × 10–10 m
- c)E = 3.4 eV, l = 6.6 × 10–11 m
- d)E = 6.8 eV, l = 6.6 × 10–11 m
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
An electron is in an excited state in hydrogen-like atom. It has a tot...
The potential energy of an electron is twice the kinetic energy(with negative sign) of an electron.
PE=2E
The total energy is: TE=PE+KE
−3.4=−2×3.4+KE
KE=3.4eV
Let p be the momentum of an electron and m be the mass of an electron.
E=p2/2m
p=√2mE
Now, the De-Broglie wavelength associated with an electron is:
λ=h/p
λ=h/√2mE
λ=6.6×1034/√2×9.1×10−31×(−3.4)×1.6×10−19
λ=6.6×1034/9.95×10−25
0λ=0.66×10−9
λ=6.6×10−10m
PE=2E
The total energy is: TE=PE+KE
−3.4=−2×3.4+KE
KE=3.4eV
Let p be the momentum of an electron and m be the mass of an electron.
E=p2/2m
p=√2mE
Now, the De-Broglie wavelength associated with an electron is:
λ=h/p
λ=h/√2mE
λ=6.6×1034/√2×9.1×10−31×(−3.4)×1.6×10−19
λ=6.6×1034/9.95×10−25
0λ=0.66×10−9
λ=6.6×10−10m
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An electron is in an excited state in hydrogen-like atom. It has a tot...
The total energy of an electron in a hydrogen-like atom can be given by the formula:
E = -13.6 eV / n^2
where E is the total energy, n is the principal quantum number.
If the electron is in an excited state, it means that its principal quantum number is greater than 1. Let's say the electron is in the excited state with a principal quantum number of n = 3.
Plugging in the values into the formula:
E = -13.6 eV / (3^2)
E = -13.6 eV / 9
E = -1.51 eV
Therefore, the electron has a total energy of approximately -1.51 eV in this excited state.
E = -13.6 eV / n^2
where E is the total energy, n is the principal quantum number.
If the electron is in an excited state, it means that its principal quantum number is greater than 1. Let's say the electron is in the excited state with a principal quantum number of n = 3.
Plugging in the values into the formula:
E = -13.6 eV / (3^2)
E = -13.6 eV / 9
E = -1.51 eV
Therefore, the electron has a total energy of approximately -1.51 eV in this excited state.
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An electron is in an excited state in hydrogen-like atom. It has a total energy of –3.4 eV. If the kinetic energy of the electron is E and its de-Broglie wavelength is l, thena)E = 6.8 eV, l = 6.6 × 10–10mb)E = 3.4 eV, l = 6.6 × 10–10mc)E = 3.4 eV, l = 6.6 × 10–11md)E = 6.8 eV, l = 6.6 × 10–11mCorrect answer is option 'B'. Can you explain this answer? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about An electron is in an excited state in hydrogen-like atom. It has a total energy of –3.4 eV. If the kinetic energy of the electron is E and its de-Broglie wavelength is l, thena)E = 6.8 eV, l = 6.6 × 10–10mb)E = 3.4 eV, l = 6.6 × 10–10mc)E = 3.4 eV, l = 6.6 × 10–11md)E = 6.8 eV, l = 6.6 × 10–11mCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An electron is in an excited state in hydrogen-like atom. It has a total energy of –3.4 eV. If the kinetic energy of the electron is E and its de-Broglie wavelength is l, thena)E = 6.8 eV, l = 6.6 × 10–10mb)E = 3.4 eV, l = 6.6 × 10–10mc)E = 3.4 eV, l = 6.6 × 10–11md)E = 6.8 eV, l = 6.6 × 10–11mCorrect answer is option 'B'. Can you explain this answer?.
An electron is in an excited state in hydrogen-like atom. It has a total energy of –3.4 eV. If the kinetic energy of the electron is E and its de-Broglie wavelength is l, thena)E = 6.8 eV, l = 6.6 × 10–10mb)E = 3.4 eV, l = 6.6 × 10–10mc)E = 3.4 eV, l = 6.6 × 10–11md)E = 6.8 eV, l = 6.6 × 10–11mCorrect answer is option 'B'. Can you explain this answer? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about An electron is in an excited state in hydrogen-like atom. It has a total energy of –3.4 eV. If the kinetic energy of the electron is E and its de-Broglie wavelength is l, thena)E = 6.8 eV, l = 6.6 × 10–10mb)E = 3.4 eV, l = 6.6 × 10–10mc)E = 3.4 eV, l = 6.6 × 10–11md)E = 6.8 eV, l = 6.6 × 10–11mCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An electron is in an excited state in hydrogen-like atom. It has a total energy of –3.4 eV. If the kinetic energy of the electron is E and its de-Broglie wavelength is l, thena)E = 6.8 eV, l = 6.6 × 10–10mb)E = 3.4 eV, l = 6.6 × 10–10mc)E = 3.4 eV, l = 6.6 × 10–11md)E = 6.8 eV, l = 6.6 × 10–11mCorrect answer is option 'B'. Can you explain this answer?.
Solutions for An electron is in an excited state in hydrogen-like atom. It has a total energy of –3.4 eV. If the kinetic energy of the electron is E and its de-Broglie wavelength is l, thena)E = 6.8 eV, l = 6.6 × 10–10mb)E = 3.4 eV, l = 6.6 × 10–10mc)E = 3.4 eV, l = 6.6 × 10–11md)E = 6.8 eV, l = 6.6 × 10–11mCorrect answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 12.
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Here you can find the meaning of An electron is in an excited state in hydrogen-like atom. It has a total energy of –3.4 eV. If the kinetic energy of the electron is E and its de-Broglie wavelength is l, thena)E = 6.8 eV, l = 6.6 × 10–10mb)E = 3.4 eV, l = 6.6 × 10–10mc)E = 3.4 eV, l = 6.6 × 10–11md)E = 6.8 eV, l = 6.6 × 10–11mCorrect answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
An electron is in an excited state in hydrogen-like atom. It has a total energy of –3.4 eV. If the kinetic energy of the electron is E and its de-Broglie wavelength is l, thena)E = 6.8 eV, l = 6.6 × 10–10mb)E = 3.4 eV, l = 6.6 × 10–10mc)E = 3.4 eV, l = 6.6 × 10–11md)E = 6.8 eV, l = 6.6 × 10–11mCorrect answer is option 'B'. Can you explain this answer?, a detailed solution for An electron is in an excited state in hydrogen-like atom. It has a total energy of –3.4 eV. If the kinetic energy of the electron is E and its de-Broglie wavelength is l, thena)E = 6.8 eV, l = 6.6 × 10–10mb)E = 3.4 eV, l = 6.6 × 10–10mc)E = 3.4 eV, l = 6.6 × 10–11md)E = 6.8 eV, l = 6.6 × 10–11mCorrect answer is option 'B'. Can you explain this answer? has been provided alongside types of An electron is in an excited state in hydrogen-like atom. It has a total energy of –3.4 eV. If the kinetic energy of the electron is E and its de-Broglie wavelength is l, thena)E = 6.8 eV, l = 6.6 × 10–10mb)E = 3.4 eV, l = 6.6 × 10–10mc)E = 3.4 eV, l = 6.6 × 10–11md)E = 6.8 eV, l = 6.6 × 10–11mCorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice An electron is in an excited state in hydrogen-like atom. It has a total energy of –3.4 eV. If the kinetic energy of the electron is E and its de-Broglie wavelength is l, thena)E = 6.8 eV, l = 6.6 × 10–10mb)E = 3.4 eV, l = 6.6 × 10–10mc)E = 3.4 eV, l = 6.6 × 10–11md)E = 6.8 eV, l = 6.6 × 10–11mCorrect answer is option 'B'. Can you explain this answer? tests, examples and also practice Class 12 tests.
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