356564
A photon is emitted in transition from \(n=4\) to \(n=1\) level in hydrogen atom. The corresponding wavelength for this transition is (given, \(h = 4 \times {10^{ - 15}}eV - s\))
1 \(974\;nm\)
2 \(99.3\;nm\)
3 \(941\;nm\)
4 \(94.1\;nm\)
Explanation:
Energy of emitted photon is given by, \(\Delta E=h v=\dfrac{h c}{\lambda}\) \(E_{1}-E_{2}=\dfrac{h c}{\lambda}\) As, \(E=\dfrac{-13.6 Z^{2}}{n^{2}} \quad[Z=1\) for hydrogen atom \(]\) \( \Rightarrow \frac{{ - 13.6{{(1)}^2}}}{{{{(4)}^2}}} - \left( {\frac{{13.6{{(1)}^2}}}{{{{(1)}^2}}}} \right)\) \( = \frac{{\left( {4 \times {{10}^{ - 15}}} \right)\left( {3 \times {{10}^8}} \right)}}{\lambda }\) \( \Rightarrow \frac{1}{\lambda } = 1.0625 \times {10^7}\;{m^{ - 1}}\) \(\lambda = 0.941 \times {10^{ - 7}}\;m\) \( \Rightarrow \lambda = 94.1\;nm\). So , correct option is (4).
JEE - 2023
PHXII12:ATOMS
356565
Match the Column-I and Column-II Column I Column II A \(n = 5\,\,to\,\,n = 2\) P Lyman series B \(n = 8\,\,to\,\,n = 4\) Q Brackett series C \(n = 3\,\,to\,\,n = 1\) R Paschen D \(n = 4\,\,to\,\,n = 3\) S Balmer
1 A-Q, B-R, C-P, D-S
2 A-S, B-Q, C-P, D-R
3 A-R, B-Q, C-S, D-P,S
4 A-P,R, B-S, C-R, D-P
Explanation:
Option (2) is correct.
PHXII12:ATOMS
356566
The following diagram indicates the energy levels of a certain atom when the system moves from \(2\,E\) level to \(E\), a photon of wavelength \(\lambda\) is emitted. The wavelength of photon produced during its transistion from \(\dfrac{4 E}{3}\) level to \(E\) is:
356567
When the electron in hydrogen atom jumps from fourth Bohr orbit to second Bohr orbit, one gets the
1 Second line of Balmer series
2 First line of Balmer series
3 First line of Pfund series
4 Second line of Paschen series
Explanation:
The wavelength of line in case of Balmer series is given by \(\frac{1}{\lambda } = R\left( {\frac{1}{{{2^2}}} - \frac{1}{{{n^2}}}} \right)\) where \(n = 3,4,5\) and \(R = \) Rydberg constant So, for Balmer series, the transition takes from third orbit to second for first line spectrum, fourth orbit to second for second line spectrum and so on. Hence, given transition represents second line of Balmer series
MHTCET - 2019
PHXII12:ATOMS
356568
If \({\lambda _1}\,and\,{\lambda _2}\) are the wavelength of the Second member of the Lyman and Paschen series respectively, then \({\lambda _1}\,:{\lambda _2}\) is
1 \(1:3\)
2 \(1:30\)
3 \(7:50\)
4 \(18:225\)
Explanation:
For second line of Lyman series,\({n_1} = 1\) and \({n_2} = 3\) \(\therefore \quad \frac{1}{{{\lambda _1}}} = R\left( {\frac{1}{{{1^2}}} - \frac{1}{{{3^2}}}} \right) = R\left( {1 - \frac{1}{9}} \right) = \frac{{8R}}{9}\) For second line of paschen series, \({n_1} = 3\) and \({n_2} = 5\) \(\therefore \quad \frac{1}{{{\lambda _2}}} = R\left( {\frac{1}{{{3^2}}} - \frac{1}{{{5^2}}}} \right) = R\left( {\frac{1}{9} - \frac{1}{{25}}} \right) = \frac{{16R}}{{225}}\) \(\therefore \quad \frac{{{\lambda _1}}}{{{\lambda _2}}} = \frac{{16R}}{{225}} \times \frac{9}{{8R}} = \frac{{18}}{{225}}\) \({\lambda _1}:{\lambda _2} = 18:225\)
356564
A photon is emitted in transition from \(n=4\) to \(n=1\) level in hydrogen atom. The corresponding wavelength for this transition is (given, \(h = 4 \times {10^{ - 15}}eV - s\))
1 \(974\;nm\)
2 \(99.3\;nm\)
3 \(941\;nm\)
4 \(94.1\;nm\)
Explanation:
Energy of emitted photon is given by, \(\Delta E=h v=\dfrac{h c}{\lambda}\) \(E_{1}-E_{2}=\dfrac{h c}{\lambda}\) As, \(E=\dfrac{-13.6 Z^{2}}{n^{2}} \quad[Z=1\) for hydrogen atom \(]\) \( \Rightarrow \frac{{ - 13.6{{(1)}^2}}}{{{{(4)}^2}}} - \left( {\frac{{13.6{{(1)}^2}}}{{{{(1)}^2}}}} \right)\) \( = \frac{{\left( {4 \times {{10}^{ - 15}}} \right)\left( {3 \times {{10}^8}} \right)}}{\lambda }\) \( \Rightarrow \frac{1}{\lambda } = 1.0625 \times {10^7}\;{m^{ - 1}}\) \(\lambda = 0.941 \times {10^{ - 7}}\;m\) \( \Rightarrow \lambda = 94.1\;nm\). So , correct option is (4).
JEE - 2023
PHXII12:ATOMS
356565
Match the Column-I and Column-II Column I Column II A \(n = 5\,\,to\,\,n = 2\) P Lyman series B \(n = 8\,\,to\,\,n = 4\) Q Brackett series C \(n = 3\,\,to\,\,n = 1\) R Paschen D \(n = 4\,\,to\,\,n = 3\) S Balmer
1 A-Q, B-R, C-P, D-S
2 A-S, B-Q, C-P, D-R
3 A-R, B-Q, C-S, D-P,S
4 A-P,R, B-S, C-R, D-P
Explanation:
Option (2) is correct.
PHXII12:ATOMS
356566
The following diagram indicates the energy levels of a certain atom when the system moves from \(2\,E\) level to \(E\), a photon of wavelength \(\lambda\) is emitted. The wavelength of photon produced during its transistion from \(\dfrac{4 E}{3}\) level to \(E\) is:
356567
When the electron in hydrogen atom jumps from fourth Bohr orbit to second Bohr orbit, one gets the
1 Second line of Balmer series
2 First line of Balmer series
3 First line of Pfund series
4 Second line of Paschen series
Explanation:
The wavelength of line in case of Balmer series is given by \(\frac{1}{\lambda } = R\left( {\frac{1}{{{2^2}}} - \frac{1}{{{n^2}}}} \right)\) where \(n = 3,4,5\) and \(R = \) Rydberg constant So, for Balmer series, the transition takes from third orbit to second for first line spectrum, fourth orbit to second for second line spectrum and so on. Hence, given transition represents second line of Balmer series
MHTCET - 2019
PHXII12:ATOMS
356568
If \({\lambda _1}\,and\,{\lambda _2}\) are the wavelength of the Second member of the Lyman and Paschen series respectively, then \({\lambda _1}\,:{\lambda _2}\) is
1 \(1:3\)
2 \(1:30\)
3 \(7:50\)
4 \(18:225\)
Explanation:
For second line of Lyman series,\({n_1} = 1\) and \({n_2} = 3\) \(\therefore \quad \frac{1}{{{\lambda _1}}} = R\left( {\frac{1}{{{1^2}}} - \frac{1}{{{3^2}}}} \right) = R\left( {1 - \frac{1}{9}} \right) = \frac{{8R}}{9}\) For second line of paschen series, \({n_1} = 3\) and \({n_2} = 5\) \(\therefore \quad \frac{1}{{{\lambda _2}}} = R\left( {\frac{1}{{{3^2}}} - \frac{1}{{{5^2}}}} \right) = R\left( {\frac{1}{9} - \frac{1}{{25}}} \right) = \frac{{16R}}{{225}}\) \(\therefore \quad \frac{{{\lambda _1}}}{{{\lambda _2}}} = \frac{{16R}}{{225}} \times \frac{9}{{8R}} = \frac{{18}}{{225}}\) \({\lambda _1}:{\lambda _2} = 18:225\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII12:ATOMS
356564
A photon is emitted in transition from \(n=4\) to \(n=1\) level in hydrogen atom. The corresponding wavelength for this transition is (given, \(h = 4 \times {10^{ - 15}}eV - s\))
1 \(974\;nm\)
2 \(99.3\;nm\)
3 \(941\;nm\)
4 \(94.1\;nm\)
Explanation:
Energy of emitted photon is given by, \(\Delta E=h v=\dfrac{h c}{\lambda}\) \(E_{1}-E_{2}=\dfrac{h c}{\lambda}\) As, \(E=\dfrac{-13.6 Z^{2}}{n^{2}} \quad[Z=1\) for hydrogen atom \(]\) \( \Rightarrow \frac{{ - 13.6{{(1)}^2}}}{{{{(4)}^2}}} - \left( {\frac{{13.6{{(1)}^2}}}{{{{(1)}^2}}}} \right)\) \( = \frac{{\left( {4 \times {{10}^{ - 15}}} \right)\left( {3 \times {{10}^8}} \right)}}{\lambda }\) \( \Rightarrow \frac{1}{\lambda } = 1.0625 \times {10^7}\;{m^{ - 1}}\) \(\lambda = 0.941 \times {10^{ - 7}}\;m\) \( \Rightarrow \lambda = 94.1\;nm\). So , correct option is (4).
JEE - 2023
PHXII12:ATOMS
356565
Match the Column-I and Column-II Column I Column II A \(n = 5\,\,to\,\,n = 2\) P Lyman series B \(n = 8\,\,to\,\,n = 4\) Q Brackett series C \(n = 3\,\,to\,\,n = 1\) R Paschen D \(n = 4\,\,to\,\,n = 3\) S Balmer
1 A-Q, B-R, C-P, D-S
2 A-S, B-Q, C-P, D-R
3 A-R, B-Q, C-S, D-P,S
4 A-P,R, B-S, C-R, D-P
Explanation:
Option (2) is correct.
PHXII12:ATOMS
356566
The following diagram indicates the energy levels of a certain atom when the system moves from \(2\,E\) level to \(E\), a photon of wavelength \(\lambda\) is emitted. The wavelength of photon produced during its transistion from \(\dfrac{4 E}{3}\) level to \(E\) is:
356567
When the electron in hydrogen atom jumps from fourth Bohr orbit to second Bohr orbit, one gets the
1 Second line of Balmer series
2 First line of Balmer series
3 First line of Pfund series
4 Second line of Paschen series
Explanation:
The wavelength of line in case of Balmer series is given by \(\frac{1}{\lambda } = R\left( {\frac{1}{{{2^2}}} - \frac{1}{{{n^2}}}} \right)\) where \(n = 3,4,5\) and \(R = \) Rydberg constant So, for Balmer series, the transition takes from third orbit to second for first line spectrum, fourth orbit to second for second line spectrum and so on. Hence, given transition represents second line of Balmer series
MHTCET - 2019
PHXII12:ATOMS
356568
If \({\lambda _1}\,and\,{\lambda _2}\) are the wavelength of the Second member of the Lyman and Paschen series respectively, then \({\lambda _1}\,:{\lambda _2}\) is
1 \(1:3\)
2 \(1:30\)
3 \(7:50\)
4 \(18:225\)
Explanation:
For second line of Lyman series,\({n_1} = 1\) and \({n_2} = 3\) \(\therefore \quad \frac{1}{{{\lambda _1}}} = R\left( {\frac{1}{{{1^2}}} - \frac{1}{{{3^2}}}} \right) = R\left( {1 - \frac{1}{9}} \right) = \frac{{8R}}{9}\) For second line of paschen series, \({n_1} = 3\) and \({n_2} = 5\) \(\therefore \quad \frac{1}{{{\lambda _2}}} = R\left( {\frac{1}{{{3^2}}} - \frac{1}{{{5^2}}}} \right) = R\left( {\frac{1}{9} - \frac{1}{{25}}} \right) = \frac{{16R}}{{225}}\) \(\therefore \quad \frac{{{\lambda _1}}}{{{\lambda _2}}} = \frac{{16R}}{{225}} \times \frac{9}{{8R}} = \frac{{18}}{{225}}\) \({\lambda _1}:{\lambda _2} = 18:225\)
356564
A photon is emitted in transition from \(n=4\) to \(n=1\) level in hydrogen atom. The corresponding wavelength for this transition is (given, \(h = 4 \times {10^{ - 15}}eV - s\))
1 \(974\;nm\)
2 \(99.3\;nm\)
3 \(941\;nm\)
4 \(94.1\;nm\)
Explanation:
Energy of emitted photon is given by, \(\Delta E=h v=\dfrac{h c}{\lambda}\) \(E_{1}-E_{2}=\dfrac{h c}{\lambda}\) As, \(E=\dfrac{-13.6 Z^{2}}{n^{2}} \quad[Z=1\) for hydrogen atom \(]\) \( \Rightarrow \frac{{ - 13.6{{(1)}^2}}}{{{{(4)}^2}}} - \left( {\frac{{13.6{{(1)}^2}}}{{{{(1)}^2}}}} \right)\) \( = \frac{{\left( {4 \times {{10}^{ - 15}}} \right)\left( {3 \times {{10}^8}} \right)}}{\lambda }\) \( \Rightarrow \frac{1}{\lambda } = 1.0625 \times {10^7}\;{m^{ - 1}}\) \(\lambda = 0.941 \times {10^{ - 7}}\;m\) \( \Rightarrow \lambda = 94.1\;nm\). So , correct option is (4).
JEE - 2023
PHXII12:ATOMS
356565
Match the Column-I and Column-II Column I Column II A \(n = 5\,\,to\,\,n = 2\) P Lyman series B \(n = 8\,\,to\,\,n = 4\) Q Brackett series C \(n = 3\,\,to\,\,n = 1\) R Paschen D \(n = 4\,\,to\,\,n = 3\) S Balmer
1 A-Q, B-R, C-P, D-S
2 A-S, B-Q, C-P, D-R
3 A-R, B-Q, C-S, D-P,S
4 A-P,R, B-S, C-R, D-P
Explanation:
Option (2) is correct.
PHXII12:ATOMS
356566
The following diagram indicates the energy levels of a certain atom when the system moves from \(2\,E\) level to \(E\), a photon of wavelength \(\lambda\) is emitted. The wavelength of photon produced during its transistion from \(\dfrac{4 E}{3}\) level to \(E\) is:
356567
When the electron in hydrogen atom jumps from fourth Bohr orbit to second Bohr orbit, one gets the
1 Second line of Balmer series
2 First line of Balmer series
3 First line of Pfund series
4 Second line of Paschen series
Explanation:
The wavelength of line in case of Balmer series is given by \(\frac{1}{\lambda } = R\left( {\frac{1}{{{2^2}}} - \frac{1}{{{n^2}}}} \right)\) where \(n = 3,4,5\) and \(R = \) Rydberg constant So, for Balmer series, the transition takes from third orbit to second for first line spectrum, fourth orbit to second for second line spectrum and so on. Hence, given transition represents second line of Balmer series
MHTCET - 2019
PHXII12:ATOMS
356568
If \({\lambda _1}\,and\,{\lambda _2}\) are the wavelength of the Second member of the Lyman and Paschen series respectively, then \({\lambda _1}\,:{\lambda _2}\) is
1 \(1:3\)
2 \(1:30\)
3 \(7:50\)
4 \(18:225\)
Explanation:
For second line of Lyman series,\({n_1} = 1\) and \({n_2} = 3\) \(\therefore \quad \frac{1}{{{\lambda _1}}} = R\left( {\frac{1}{{{1^2}}} - \frac{1}{{{3^2}}}} \right) = R\left( {1 - \frac{1}{9}} \right) = \frac{{8R}}{9}\) For second line of paschen series, \({n_1} = 3\) and \({n_2} = 5\) \(\therefore \quad \frac{1}{{{\lambda _2}}} = R\left( {\frac{1}{{{3^2}}} - \frac{1}{{{5^2}}}} \right) = R\left( {\frac{1}{9} - \frac{1}{{25}}} \right) = \frac{{16R}}{{225}}\) \(\therefore \quad \frac{{{\lambda _1}}}{{{\lambda _2}}} = \frac{{16R}}{{225}} \times \frac{9}{{8R}} = \frac{{18}}{{225}}\) \({\lambda _1}:{\lambda _2} = 18:225\)
356564
A photon is emitted in transition from \(n=4\) to \(n=1\) level in hydrogen atom. The corresponding wavelength for this transition is (given, \(h = 4 \times {10^{ - 15}}eV - s\))
1 \(974\;nm\)
2 \(99.3\;nm\)
3 \(941\;nm\)
4 \(94.1\;nm\)
Explanation:
Energy of emitted photon is given by, \(\Delta E=h v=\dfrac{h c}{\lambda}\) \(E_{1}-E_{2}=\dfrac{h c}{\lambda}\) As, \(E=\dfrac{-13.6 Z^{2}}{n^{2}} \quad[Z=1\) for hydrogen atom \(]\) \( \Rightarrow \frac{{ - 13.6{{(1)}^2}}}{{{{(4)}^2}}} - \left( {\frac{{13.6{{(1)}^2}}}{{{{(1)}^2}}}} \right)\) \( = \frac{{\left( {4 \times {{10}^{ - 15}}} \right)\left( {3 \times {{10}^8}} \right)}}{\lambda }\) \( \Rightarrow \frac{1}{\lambda } = 1.0625 \times {10^7}\;{m^{ - 1}}\) \(\lambda = 0.941 \times {10^{ - 7}}\;m\) \( \Rightarrow \lambda = 94.1\;nm\). So , correct option is (4).
JEE - 2023
PHXII12:ATOMS
356565
Match the Column-I and Column-II Column I Column II A \(n = 5\,\,to\,\,n = 2\) P Lyman series B \(n = 8\,\,to\,\,n = 4\) Q Brackett series C \(n = 3\,\,to\,\,n = 1\) R Paschen D \(n = 4\,\,to\,\,n = 3\) S Balmer
1 A-Q, B-R, C-P, D-S
2 A-S, B-Q, C-P, D-R
3 A-R, B-Q, C-S, D-P,S
4 A-P,R, B-S, C-R, D-P
Explanation:
Option (2) is correct.
PHXII12:ATOMS
356566
The following diagram indicates the energy levels of a certain atom when the system moves from \(2\,E\) level to \(E\), a photon of wavelength \(\lambda\) is emitted. The wavelength of photon produced during its transistion from \(\dfrac{4 E}{3}\) level to \(E\) is:
356567
When the electron in hydrogen atom jumps from fourth Bohr orbit to second Bohr orbit, one gets the
1 Second line of Balmer series
2 First line of Balmer series
3 First line of Pfund series
4 Second line of Paschen series
Explanation:
The wavelength of line in case of Balmer series is given by \(\frac{1}{\lambda } = R\left( {\frac{1}{{{2^2}}} - \frac{1}{{{n^2}}}} \right)\) where \(n = 3,4,5\) and \(R = \) Rydberg constant So, for Balmer series, the transition takes from third orbit to second for first line spectrum, fourth orbit to second for second line spectrum and so on. Hence, given transition represents second line of Balmer series
MHTCET - 2019
PHXII12:ATOMS
356568
If \({\lambda _1}\,and\,{\lambda _2}\) are the wavelength of the Second member of the Lyman and Paschen series respectively, then \({\lambda _1}\,:{\lambda _2}\) is
1 \(1:3\)
2 \(1:30\)
3 \(7:50\)
4 \(18:225\)
Explanation:
For second line of Lyman series,\({n_1} = 1\) and \({n_2} = 3\) \(\therefore \quad \frac{1}{{{\lambda _1}}} = R\left( {\frac{1}{{{1^2}}} - \frac{1}{{{3^2}}}} \right) = R\left( {1 - \frac{1}{9}} \right) = \frac{{8R}}{9}\) For second line of paschen series, \({n_1} = 3\) and \({n_2} = 5\) \(\therefore \quad \frac{1}{{{\lambda _2}}} = R\left( {\frac{1}{{{3^2}}} - \frac{1}{{{5^2}}}} \right) = R\left( {\frac{1}{9} - \frac{1}{{25}}} \right) = \frac{{16R}}{{225}}\) \(\therefore \quad \frac{{{\lambda _1}}}{{{\lambda _2}}} = \frac{{16R}}{{225}} \times \frac{9}{{8R}} = \frac{{18}}{{225}}\) \({\lambda _1}:{\lambda _2} = 18:225\)
