Determine the maximum number of emission lines produced when an electron in a hydrogen atom transitions from the n = 6 energy level to the ground state :
1. 30
2. 21
3. 15
4. 28
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NEET MCQ Books for XIth & XIIth Physics, Chemistry & BiologyThe ionization energy for H atom in the ground state is \(2.18 \times10^{-18}~ \mathrm J.\) The process energy requirements will be: \(( He^+(g) \rightarrow He^{2+}(g) + e^- )\)
1. | \(8.72 \times10^{-18}~\mathrm J\) | 2. | \(7.54 \times10^{-18}~\mathrm J\) |
3. | \(5.67 \times10^{-17}~\mathrm J\) | 4. | \(2.18 \times10^{-17}~\mathrm J\) |
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NEET MCQ Books for XIth & XIIth Physics, Chemistry & BiologyThe wavelength of light emitted when the electron in a H atom undergoes the transition from an energy level with n = 4 to an energy level with n = 2, is :
1. 586 mm
2. 486 nm
3. 523 nm
4. 416 pm
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NEET MCQ Books for XIth & XIIth Physics, Chemistry & BiologyThe energy associated with the fifth orbit of a hydrogen atom is :
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NEET MCQ Books for XIth & XIIth Physics, Chemistry & BiologyThe wave number for the longest wavelength transition in the Balmer series of atomic hydrogen would be :
\(1 . 1 . 52 \times 10^{6} m^{- 1}\)
\(2 . 3 . 14 \times 10^{6} \left(cm\right)^{- 1}\)
\(3 . 15 . 2 \times 10^{6} m^{- 1}\)
\(4 . 1 . 52 \times 10^{6} \left(cm\right)^{- 1}\)
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NEET MCQ Books for XIth & XIIth Physics, Chemistry & BiologyThe energy of an electron in an H - atom is given by \(E_n=(-2.18 \times10^{-18})/n^2~\mathrm J.\) The shortest wavelength of light that can be used to remove an electron completely from \(n = 2\) orbit will be:
1. | \(3647~\mathring{\mathrm A}\) | 2. | \(5132~\mathring{\mathrm A}\) |
3. | \(3017~\mathring{\mathrm A}\) | 4. | None of these |
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NEET MCQ Books for XIth & XIIth Physics, Chemistry & BiologyEmission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represented as v = 3.29 × 1015 (Hz) . The value of n if the transition is observed at 1285 nm is :
1. | 6 | 2. | 5 |
3. | 8 | 4. | 9 |
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NEET MCQ Books for XIth & XIIth Physics, Chemistry & BiologyThe transition in the hydrogen spectrum that would have the same wavelength as Balmer transition from n = 4 to n = 2 of He+ spectrum is :
1.
2. = 3 to n1 = 2
3. = 3 to n1 = 1
4. = 2 to n1 = 1
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NEET MCQ Books for XIth & XIIth Physics, Chemistry & Biology