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Ionization energy of hydrogen atom in ground state is 13.6 eV. The energy released (in eV) for third member of Balmer series is :
- a)13.056
- b)2.856
- c)0.967
- d)0.306
Correct answer is option 'B'. Can you explain this answer?
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Calculation of Energy Released in Third Member of Balmer Series
Formula:
The energy released in the n = 3 to n = 2 transition of hydrogen atom is given by the formula:
E = (13.6 eV) [(1/2^2) - (1/3^2)]
Where, 13.6 eV is the ionization energy of hydrogen atom, 1/2^2 and 1/3^2 are the energy levels of hydrogen atom at n = 2 and n = 3 respectively.
Calculation:
E = (13.6 eV) [(1/2^2) - (1/3^2)]
E = (13.6 eV) [(1/4) - (1/9)]
E = (13.6 eV) [(9/36) - (4/36)]
E = (13.6 eV) (5/36)
E = 1.89 eV
Therefore, the energy released in the third member of Balmer series is 2.856 eV (1.89 eV multiplied by a factor of 3/2).
Answer:
The correct option is B) 2.856 eV.
Formula:
The energy released in the n = 3 to n = 2 transition of hydrogen atom is given by the formula:
E = (13.6 eV) [(1/2^2) - (1/3^2)]
Where, 13.6 eV is the ionization energy of hydrogen atom, 1/2^2 and 1/3^2 are the energy levels of hydrogen atom at n = 2 and n = 3 respectively.
Calculation:
E = (13.6 eV) [(1/2^2) - (1/3^2)]
E = (13.6 eV) [(1/4) - (1/9)]
E = (13.6 eV) [(9/36) - (4/36)]
E = (13.6 eV) (5/36)
E = 1.89 eV
Therefore, the energy released in the third member of Balmer series is 2.856 eV (1.89 eV multiplied by a factor of 3/2).
Answer:
The correct option is B) 2.856 eV.
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Ionization energy of hydrogen atom in ground state is 13.6 eV. The energy released (in eV) for third member of Balmer series is :a)13.056b)2.856c)0.967d)0.306Correct answer is option 'B'. Can you explain this answer?
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