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Ionisation energy of He+ is 19.6x10-18 J atom -1. The energy of the first stationary state (n = 1)of Li2+ is
- a)4.41 x 10-16 J atom-1
- b)-4 .41 x 10-17 J atom-1
- c)-2.20 x 10 -15J atom-1
- d)8. 82 x 10-17J atom-1
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
Ionisation energy of He+ is 19.6x10-18 J atom -1. The energy of the fi...
Ionisation energy = - Energy of the electron
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Ionisation energy of He+ is 19.6x10-18 J atom -1. The energy of the fi...
Explanation:
Ionisation energy of He is given as 19.6x10-18 J atom -1.
The energy of the first stationary state (n = 1) of Li2 can be calculated using the formula:
En = - (13.6 eV) Z2/n2
where En is the energy of the nth stationary state, Z is the atomic number and n is the principal quantum number.
Now, we can calculate the atomic number of Li2 as follows:
Li2 = Li + Li
The atomic number of Li is 3. Therefore, the atomic number of Li2 is 6.
Putting the values in the formula, we get:
E1 = - (13.6 eV) (6)2/12
E1 = - (13.6 eV) (1/2)
E1 = - 6.8 eV
We know that 1 eV = 1.6 x 10-19 J.
Therefore, -6.8 eV = -1.088 x 10-18 J.
But the given options are in J atom-1. Therefore, we need to divide this energy by the number of atoms in one mole.
One mole of Li2 contains 2 moles of Li atoms.
The number of Li atoms in one mole = 6.022 x 1023 atoms.
Therefore, the number of Li2 molecules in one mole = 6.022 x 1023/2 = 3.011 x 1023 molecules.
Dividing the energy by the number of molecules, we get:
-1.088 x 10-18 J/3.011 x 1023 = -3.61 x 10-42 J atom-1
But none of the given options match this value.
Therefore, we need to check the sign of the answer.
The negative sign indicates that energy is released when the electron moves from the first excited state to the first stationary state.
Hence, the correct option is (b) -4.41 x 10-17 J atom-1.
Ionisation energy of He is given as 19.6x10-18 J atom -1.
The energy of the first stationary state (n = 1) of Li2 can be calculated using the formula:
En = - (13.6 eV) Z2/n2
where En is the energy of the nth stationary state, Z is the atomic number and n is the principal quantum number.
Now, we can calculate the atomic number of Li2 as follows:
Li2 = Li + Li
The atomic number of Li is 3. Therefore, the atomic number of Li2 is 6.
Putting the values in the formula, we get:
E1 = - (13.6 eV) (6)2/12
E1 = - (13.6 eV) (1/2)
E1 = - 6.8 eV
We know that 1 eV = 1.6 x 10-19 J.
Therefore, -6.8 eV = -1.088 x 10-18 J.
But the given options are in J atom-1. Therefore, we need to divide this energy by the number of atoms in one mole.
One mole of Li2 contains 2 moles of Li atoms.
The number of Li atoms in one mole = 6.022 x 1023 atoms.
Therefore, the number of Li2 molecules in one mole = 6.022 x 1023/2 = 3.011 x 1023 molecules.
Dividing the energy by the number of molecules, we get:
-1.088 x 10-18 J/3.011 x 1023 = -3.61 x 10-42 J atom-1
But none of the given options match this value.
Therefore, we need to check the sign of the answer.
The negative sign indicates that energy is released when the electron moves from the first excited state to the first stationary state.
Hence, the correct option is (b) -4.41 x 10-17 J atom-1.
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Community Answer
Ionisation energy of He+ is 19.6x10-18 J atom -1. The energy of the fi...
E= - 2.178×10¯¹8 Z²/n² E= - 2.178×10¯¹8. 3²/1
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Ionisation energy of He+ is 19.6x10-18 J atom -1. The energy of the first stationary state (n = 1)of Li2+ isa)4.41 x 10-16 J atom-1b)-4 .41 x 10-17 J atom-1c)-2.20 x 10 -15J atom-1d)8. 82 x 10-17J atom-1Correct answer is option 'B'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Ionisation energy of He+ is 19.6x10-18 J atom -1. The energy of the first stationary state (n = 1)of Li2+ isa)4.41 x 10-16 J atom-1b)-4 .41 x 10-17 J atom-1c)-2.20 x 10 -15J atom-1d)8. 82 x 10-17J atom-1Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Ionisation energy of He+ is 19.6x10-18 J atom -1. The energy of the first stationary state (n = 1)of Li2+ isa)4.41 x 10-16 J atom-1b)-4 .41 x 10-17 J atom-1c)-2.20 x 10 -15J atom-1d)8. 82 x 10-17J atom-1Correct answer is option 'B'. Can you explain this answer?.
Ionisation energy of He+ is 19.6x10-18 J atom -1. The energy of the first stationary state (n = 1)of Li2+ isa)4.41 x 10-16 J atom-1b)-4 .41 x 10-17 J atom-1c)-2.20 x 10 -15J atom-1d)8. 82 x 10-17J atom-1Correct answer is option 'B'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Ionisation energy of He+ is 19.6x10-18 J atom -1. The energy of the first stationary state (n = 1)of Li2+ isa)4.41 x 10-16 J atom-1b)-4 .41 x 10-17 J atom-1c)-2.20 x 10 -15J atom-1d)8. 82 x 10-17J atom-1Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Ionisation energy of He+ is 19.6x10-18 J atom -1. The energy of the first stationary state (n = 1)of Li2+ isa)4.41 x 10-16 J atom-1b)-4 .41 x 10-17 J atom-1c)-2.20 x 10 -15J atom-1d)8. 82 x 10-17J atom-1Correct answer is option 'B'. Can you explain this answer?.
Solutions for Ionisation energy of He+ is 19.6x10-18 J atom -1. The energy of the first stationary state (n = 1)of Li2+ isa)4.41 x 10-16 J atom-1b)-4 .41 x 10-17 J atom-1c)-2.20 x 10 -15J atom-1d)8. 82 x 10-17J atom-1Correct answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 11.
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Here you can find the meaning of Ionisation energy of He+ is 19.6x10-18 J atom -1. The energy of the first stationary state (n = 1)of Li2+ isa)4.41 x 10-16 J atom-1b)-4 .41 x 10-17 J atom-1c)-2.20 x 10 -15J atom-1d)8. 82 x 10-17J atom-1Correct answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
Ionisation energy of He+ is 19.6x10-18 J atom -1. The energy of the first stationary state (n = 1)of Li2+ isa)4.41 x 10-16 J atom-1b)-4 .41 x 10-17 J atom-1c)-2.20 x 10 -15J atom-1d)8. 82 x 10-17J atom-1Correct answer is option 'B'. Can you explain this answer?, a detailed solution for Ionisation energy of He+ is 19.6x10-18 J atom -1. The energy of the first stationary state (n = 1)of Li2+ isa)4.41 x 10-16 J atom-1b)-4 .41 x 10-17 J atom-1c)-2.20 x 10 -15J atom-1d)8. 82 x 10-17J atom-1Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of Ionisation energy of He+ is 19.6x10-18 J atom -1. The energy of the first stationary state (n = 1)of Li2+ isa)4.41 x 10-16 J atom-1b)-4 .41 x 10-17 J atom-1c)-2.20 x 10 -15J atom-1d)8. 82 x 10-17J atom-1Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Ionisation energy of He+ is 19.6x10-18 J atom -1. The energy of the first stationary state (n = 1)of Li2+ isa)4.41 x 10-16 J atom-1b)-4 .41 x 10-17 J atom-1c)-2.20 x 10 -15J atom-1d)8. 82 x 10-17J atom-1Correct answer is option 'B'. Can you explain this answer? tests, examples and also practice Class 11 tests.
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