356637
Atomic hydrogen is excited to the \({n^{th}}\) energy level.The maximum number of spectral lines which it can emit while returning to the ground state, is:
1 \(\frac{1}{2}n(n - 1)\)
2 \(\frac{1}{2}n(n + 1)\)
3 \(n(n - 1)\)
4 \(n(n + 1)\)
Explanation:
Conceptual Question
PHXII12:ATOMS
356638
Given the longest wavelength in Lyman series as \(1240A^\circ \), the highest frequency emitted in Balmer series is
1 \(8 \times {10^{14}}Hz\)
2 \(8 \times {10^{12}}Hz\)
3 \(8 \times {10^{10}}Hz\)
4 \(8 \times {10^3}Hz\)
Explanation:
The wavelength of different spectral lines of Lyman series is given by \(\frac{1}{\lambda } = R\left[ {\frac{1}{{{1^2}}} - \frac{1}{{n_2^2}}} \right]\) where \({n_2} = 2,3,4,....\) For longest wavelength of Lyman series \({n_2} = 2\) \(\therefore \frac{1}{{{\lambda _L}}} = R\left[ {\frac{1}{{{1^2}}} - \frac{1}{{{2^2}}}} \right] = R\left[ {\frac{3}{4}} \right]\) Highest frequency of Balmer series correspond to the smallest wavelength of Balmer series. The wavelength of different spectral lines of Balmer series is given by \(\frac{1}{\lambda } = R\left[ {\frac{1}{{{2^2}}} - \frac{1}{{n_2^2}}} \right]{\rm{where}}\,\,{n_2} = 3,4,5,...\) For smallest wavelength of Balmer series \({n_2} = \infty \) \(\therefore \frac{1}{{{\lambda _S}}} = R\left[ {\frac{1}{{{2^2}}} - \frac{1}{{{\infty ^2}}}} \right] = \frac{R}{4}\) \(\therefore \frac{{{\lambda _S}}}{{{\lambda _L}}} = \frac{{3R}}{4} \times \frac{4}{R} = 3\) or \({\lambda _S} = 3{\lambda _L} = 3 \times 1240A^\circ \) Highest frequency emitted in Balmer series is \({v_H} = \frac{c}{{{\lambda _S}}} = \frac{{3 \times {{10}^8}}}{{3 \times 1240 \times {{10}^{ - 10}}}} = 8 \times {10^{14}}Hz\)
PHXII12:ATOMS
356639
The shortest wavelenght of Paschen, Balmer and Lyman series are in the ratio:
356640
An electron of a stationary hydrogen atom passes from the fifth energy level to the ground level. The velocity that the atom acquired as a result of photon emission will be [\(m = \) mass of atom]
356637
Atomic hydrogen is excited to the \({n^{th}}\) energy level.The maximum number of spectral lines which it can emit while returning to the ground state, is:
1 \(\frac{1}{2}n(n - 1)\)
2 \(\frac{1}{2}n(n + 1)\)
3 \(n(n - 1)\)
4 \(n(n + 1)\)
Explanation:
Conceptual Question
PHXII12:ATOMS
356638
Given the longest wavelength in Lyman series as \(1240A^\circ \), the highest frequency emitted in Balmer series is
1 \(8 \times {10^{14}}Hz\)
2 \(8 \times {10^{12}}Hz\)
3 \(8 \times {10^{10}}Hz\)
4 \(8 \times {10^3}Hz\)
Explanation:
The wavelength of different spectral lines of Lyman series is given by \(\frac{1}{\lambda } = R\left[ {\frac{1}{{{1^2}}} - \frac{1}{{n_2^2}}} \right]\) where \({n_2} = 2,3,4,....\) For longest wavelength of Lyman series \({n_2} = 2\) \(\therefore \frac{1}{{{\lambda _L}}} = R\left[ {\frac{1}{{{1^2}}} - \frac{1}{{{2^2}}}} \right] = R\left[ {\frac{3}{4}} \right]\) Highest frequency of Balmer series correspond to the smallest wavelength of Balmer series. The wavelength of different spectral lines of Balmer series is given by \(\frac{1}{\lambda } = R\left[ {\frac{1}{{{2^2}}} - \frac{1}{{n_2^2}}} \right]{\rm{where}}\,\,{n_2} = 3,4,5,...\) For smallest wavelength of Balmer series \({n_2} = \infty \) \(\therefore \frac{1}{{{\lambda _S}}} = R\left[ {\frac{1}{{{2^2}}} - \frac{1}{{{\infty ^2}}}} \right] = \frac{R}{4}\) \(\therefore \frac{{{\lambda _S}}}{{{\lambda _L}}} = \frac{{3R}}{4} \times \frac{4}{R} = 3\) or \({\lambda _S} = 3{\lambda _L} = 3 \times 1240A^\circ \) Highest frequency emitted in Balmer series is \({v_H} = \frac{c}{{{\lambda _S}}} = \frac{{3 \times {{10}^8}}}{{3 \times 1240 \times {{10}^{ - 10}}}} = 8 \times {10^{14}}Hz\)
PHXII12:ATOMS
356639
The shortest wavelenght of Paschen, Balmer and Lyman series are in the ratio:
356640
An electron of a stationary hydrogen atom passes from the fifth energy level to the ground level. The velocity that the atom acquired as a result of photon emission will be [\(m = \) mass of atom]
356637
Atomic hydrogen is excited to the \({n^{th}}\) energy level.The maximum number of spectral lines which it can emit while returning to the ground state, is:
1 \(\frac{1}{2}n(n - 1)\)
2 \(\frac{1}{2}n(n + 1)\)
3 \(n(n - 1)\)
4 \(n(n + 1)\)
Explanation:
Conceptual Question
PHXII12:ATOMS
356638
Given the longest wavelength in Lyman series as \(1240A^\circ \), the highest frequency emitted in Balmer series is
1 \(8 \times {10^{14}}Hz\)
2 \(8 \times {10^{12}}Hz\)
3 \(8 \times {10^{10}}Hz\)
4 \(8 \times {10^3}Hz\)
Explanation:
The wavelength of different spectral lines of Lyman series is given by \(\frac{1}{\lambda } = R\left[ {\frac{1}{{{1^2}}} - \frac{1}{{n_2^2}}} \right]\) where \({n_2} = 2,3,4,....\) For longest wavelength of Lyman series \({n_2} = 2\) \(\therefore \frac{1}{{{\lambda _L}}} = R\left[ {\frac{1}{{{1^2}}} - \frac{1}{{{2^2}}}} \right] = R\left[ {\frac{3}{4}} \right]\) Highest frequency of Balmer series correspond to the smallest wavelength of Balmer series. The wavelength of different spectral lines of Balmer series is given by \(\frac{1}{\lambda } = R\left[ {\frac{1}{{{2^2}}} - \frac{1}{{n_2^2}}} \right]{\rm{where}}\,\,{n_2} = 3,4,5,...\) For smallest wavelength of Balmer series \({n_2} = \infty \) \(\therefore \frac{1}{{{\lambda _S}}} = R\left[ {\frac{1}{{{2^2}}} - \frac{1}{{{\infty ^2}}}} \right] = \frac{R}{4}\) \(\therefore \frac{{{\lambda _S}}}{{{\lambda _L}}} = \frac{{3R}}{4} \times \frac{4}{R} = 3\) or \({\lambda _S} = 3{\lambda _L} = 3 \times 1240A^\circ \) Highest frequency emitted in Balmer series is \({v_H} = \frac{c}{{{\lambda _S}}} = \frac{{3 \times {{10}^8}}}{{3 \times 1240 \times {{10}^{ - 10}}}} = 8 \times {10^{14}}Hz\)
PHXII12:ATOMS
356639
The shortest wavelenght of Paschen, Balmer and Lyman series are in the ratio:
356640
An electron of a stationary hydrogen atom passes from the fifth energy level to the ground level. The velocity that the atom acquired as a result of photon emission will be [\(m = \) mass of atom]
356637
Atomic hydrogen is excited to the \({n^{th}}\) energy level.The maximum number of spectral lines which it can emit while returning to the ground state, is:
1 \(\frac{1}{2}n(n - 1)\)
2 \(\frac{1}{2}n(n + 1)\)
3 \(n(n - 1)\)
4 \(n(n + 1)\)
Explanation:
Conceptual Question
PHXII12:ATOMS
356638
Given the longest wavelength in Lyman series as \(1240A^\circ \), the highest frequency emitted in Balmer series is
1 \(8 \times {10^{14}}Hz\)
2 \(8 \times {10^{12}}Hz\)
3 \(8 \times {10^{10}}Hz\)
4 \(8 \times {10^3}Hz\)
Explanation:
The wavelength of different spectral lines of Lyman series is given by \(\frac{1}{\lambda } = R\left[ {\frac{1}{{{1^2}}} - \frac{1}{{n_2^2}}} \right]\) where \({n_2} = 2,3,4,....\) For longest wavelength of Lyman series \({n_2} = 2\) \(\therefore \frac{1}{{{\lambda _L}}} = R\left[ {\frac{1}{{{1^2}}} - \frac{1}{{{2^2}}}} \right] = R\left[ {\frac{3}{4}} \right]\) Highest frequency of Balmer series correspond to the smallest wavelength of Balmer series. The wavelength of different spectral lines of Balmer series is given by \(\frac{1}{\lambda } = R\left[ {\frac{1}{{{2^2}}} - \frac{1}{{n_2^2}}} \right]{\rm{where}}\,\,{n_2} = 3,4,5,...\) For smallest wavelength of Balmer series \({n_2} = \infty \) \(\therefore \frac{1}{{{\lambda _S}}} = R\left[ {\frac{1}{{{2^2}}} - \frac{1}{{{\infty ^2}}}} \right] = \frac{R}{4}\) \(\therefore \frac{{{\lambda _S}}}{{{\lambda _L}}} = \frac{{3R}}{4} \times \frac{4}{R} = 3\) or \({\lambda _S} = 3{\lambda _L} = 3 \times 1240A^\circ \) Highest frequency emitted in Balmer series is \({v_H} = \frac{c}{{{\lambda _S}}} = \frac{{3 \times {{10}^8}}}{{3 \times 1240 \times {{10}^{ - 10}}}} = 8 \times {10^{14}}Hz\)
PHXII12:ATOMS
356639
The shortest wavelenght of Paschen, Balmer and Lyman series are in the ratio:
356640
An electron of a stationary hydrogen atom passes from the fifth energy level to the ground level. The velocity that the atom acquired as a result of photon emission will be [\(m = \) mass of atom]
