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The excitataion energy of an electron from second orbit to third orbit of an atom with źe nuclear charge is 47.2 if the energy of h-atom in lowest energy state is -13.6ev what will be the value of z?
Verified Answer
The excitataion energy of an electron from second orbit to third orbit...
Assuming atom is Hydrogen like.
Energy of nth orbit is given by
-13.6Z^2 / n^2 eV
So, Energy of 2nd orbit = -3.4 Z^2 eV and
Energy of 3rd orbit = -1.51 Z^2 eV
Difference in energy between 2ed and 3rd orbit = final energy - initial energy = 1.89 Z^2 eV
So, 1.89Z^2 = 47.2 , solve for Z and round off to the nearest integer
Z= 5
Energy of nth orbit is given by
-13.6Z^2 / n^2 eV
So, Energy of 2nd orbit = -3.4 Z^2 eV and
Energy of 3rd orbit = -1.51 Z^2 eV
Difference in energy between 2ed and 3rd orbit = final energy - initial energy = 1.89 Z^2 eV
So, 1.89Z^2 = 47.2 , solve for Z and round off to the nearest integer
Z= 5
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Most Upvoted Answer
The excitataion energy of an electron from second orbit to third orbit...
Understanding Electron Excitation Energy
To find the nuclear charge (Z) in the given scenario, we can use the formula for the energy levels of a hydrogen-like atom. The energy of an electron in the n-th orbit is given by:
- E_n = - (Z^2 * 13.6 eV) / n^2
Energy Levels for the Given Orbits
- For the second orbit (n=2):
- E_2 = - (Z^2 * 13.6 eV) / 2^2 = - (Z^2 * 13.6 eV) / 4
- For the third orbit (n=3):
- E_3 = - (Z^2 * 13.6 eV) / 3^2 = - (Z^2 * 13.6 eV) / 9
Calculating Excitation Energy
- The excitation energy (ΔE) from n=2 to n=3 is given as 47.2 eV:
- ΔE = E_3 - E_2
- Substituting the expressions for E_3 and E_2:
- ΔE = [- (Z^2 * 13.6 eV) / 9] - [- (Z^2 * 13.6 eV) / 4]
Finding a Common Denominator
- The common denominator for 9 and 4 is 36:
- ΔE = (Z^2 * 13.6 eV) * [(4/36) - (9/36)]
- ΔE = (Z^2 * 13.6 eV) * [-5/36]
Setting Up the Equation
- Now, set the equation equal to the excitation energy:
- 47.2 eV = - (Z^2 * 13.6 eV * 5/36)
Solving for Z
- Rearranging gives:
- Z^2 = (47.2 * 36) / (13.6 * 5)
- This simplifies to Z^2 = 62.4, leading to:
- Z = √62.4 ≈ 7.9
Conclusion
- The nuclear charge Z is approximately 8. Thus, the atom corresponds to oxygen, which has 8 protons.
To find the nuclear charge (Z) in the given scenario, we can use the formula for the energy levels of a hydrogen-like atom. The energy of an electron in the n-th orbit is given by:
- E_n = - (Z^2 * 13.6 eV) / n^2
Energy Levels for the Given Orbits
- For the second orbit (n=2):
- E_2 = - (Z^2 * 13.6 eV) / 2^2 = - (Z^2 * 13.6 eV) / 4
- For the third orbit (n=3):
- E_3 = - (Z^2 * 13.6 eV) / 3^2 = - (Z^2 * 13.6 eV) / 9
Calculating Excitation Energy
- The excitation energy (ΔE) from n=2 to n=3 is given as 47.2 eV:
- ΔE = E_3 - E_2
- Substituting the expressions for E_3 and E_2:
- ΔE = [- (Z^2 * 13.6 eV) / 9] - [- (Z^2 * 13.6 eV) / 4]
Finding a Common Denominator
- The common denominator for 9 and 4 is 36:
- ΔE = (Z^2 * 13.6 eV) * [(4/36) - (9/36)]
- ΔE = (Z^2 * 13.6 eV) * [-5/36]
Setting Up the Equation
- Now, set the equation equal to the excitation energy:
- 47.2 eV = - (Z^2 * 13.6 eV * 5/36)
Solving for Z
- Rearranging gives:
- Z^2 = (47.2 * 36) / (13.6 * 5)
- This simplifies to Z^2 = 62.4, leading to:
- Z = √62.4 ≈ 7.9
Conclusion
- The nuclear charge Z is approximately 8. Thus, the atom corresponds to oxygen, which has 8 protons.
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The excitataion energy of an electron from second orbit to third orbit of an atom with źe nuclear charge is 47.2 if the energy of h-atom in lowest energy state is -13.6ev what will be the value of z?
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The excitataion energy of an electron from second orbit to third orbit of an atom with źe nuclear charge is 47.2 if the energy of h-atom in lowest energy state is -13.6ev what will be the value of z? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The excitataion energy of an electron from second orbit to third orbit of an atom with źe nuclear charge is 47.2 if the energy of h-atom in lowest energy state is -13.6ev what will be the value of z? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The excitataion energy of an electron from second orbit to third orbit of an atom with źe nuclear charge is 47.2 if the energy of h-atom in lowest energy state is -13.6ev what will be the value of z?.
The excitataion energy of an electron from second orbit to third orbit of an atom with źe nuclear charge is 47.2 if the energy of h-atom in lowest energy state is -13.6ev what will be the value of z? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The excitataion energy of an electron from second orbit to third orbit of an atom with źe nuclear charge is 47.2 if the energy of h-atom in lowest energy state is -13.6ev what will be the value of z? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The excitataion energy of an electron from second orbit to third orbit of an atom with źe nuclear charge is 47.2 if the energy of h-atom in lowest energy state is -13.6ev what will be the value of z?.
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