The Line Spectra of the Hydrogen Atom
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PHXII12:ATOMS

356633 Calculate the highest frequency of the emitted photon in the Paschen series of spectral lines of the hydrogen atom.

1 \(3.7 \times {10^{14}}\;Hz\)
2 \(9.1 \times {10^{15}}\;Hz\)
3 \(10.23 \times {10^{14}}\;Hz\)
4 \(29.7 \times {10^{15}}\;Hz\)
PHXII12:ATOMS

356634 In hydrogen spectrum, the shortest wavelength in the Balmer series is \(\lambda\). The shortset wavelength in the Bracket series is:

1 \(4 \lambda\)
2 \(9 \lambda\)
3 \(16 \lambda\)
4 \(2 \lambda\)
NEET - 2023
PHXII12:ATOMS

356635 Hydrogen atom from excited state comes to the ground state by emitting a photon of wavelength \(\lambda \).If \(R\) is the Rydberg constant, the principal quantum number \(n\) of the excited state is

1 \(\sqrt {\frac{{\lambda R}}{{\lambda R - 1}}} \)
2 \(\sqrt {\frac{\lambda }{{\lambda R - 1}}} \)
3 \(\sqrt {\frac{{\lambda {R^2}}}{{\lambda R - 1}}} \)
4 \(\sqrt {\frac{{\lambda R}}{{\lambda - 1}}} \)
KCET - 2009
PHXII12:ATOMS

356636 Three energy levels of hydrogen atom and the corresponding wavelength of the emitted radiation due to different electron transition are as shown. Then.
supporting img

1 \(\lambda_{2}=\lambda_{1}+\lambda_{3}\)
2 \(\lambda_{2}=\dfrac{\lambda_{1} \lambda_{3}}{\lambda_{1}+\lambda_{3}}\)
3 \(\lambda_{3}=\dfrac{\lambda_{1} \lambda_{2}}{\lambda_{1}+\lambda_{2}}\)
4 \(\lambda_{1}=\dfrac{\lambda_{2} \lambda_{3}}{\lambda_{2}+\lambda_{3}}\)
KCET - 2023
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PHXII12:ATOMS

356633 Calculate the highest frequency of the emitted photon in the Paschen series of spectral lines of the hydrogen atom.

1 \(3.7 \times {10^{14}}\;Hz\)
2 \(9.1 \times {10^{15}}\;Hz\)
3 \(10.23 \times {10^{14}}\;Hz\)
4 \(29.7 \times {10^{15}}\;Hz\)
PHXII12:ATOMS

356634 In hydrogen spectrum, the shortest wavelength in the Balmer series is \(\lambda\). The shortset wavelength in the Bracket series is:

1 \(4 \lambda\)
2 \(9 \lambda\)
3 \(16 \lambda\)
4 \(2 \lambda\)
NEET - 2023
PHXII12:ATOMS

356635 Hydrogen atom from excited state comes to the ground state by emitting a photon of wavelength \(\lambda \).If \(R\) is the Rydberg constant, the principal quantum number \(n\) of the excited state is

1 \(\sqrt {\frac{{\lambda R}}{{\lambda R - 1}}} \)
2 \(\sqrt {\frac{\lambda }{{\lambda R - 1}}} \)
3 \(\sqrt {\frac{{\lambda {R^2}}}{{\lambda R - 1}}} \)
4 \(\sqrt {\frac{{\lambda R}}{{\lambda - 1}}} \)
KCET - 2009
PHXII12:ATOMS

356636 Three energy levels of hydrogen atom and the corresponding wavelength of the emitted radiation due to different electron transition are as shown. Then.
supporting img

1 \(\lambda_{2}=\lambda_{1}+\lambda_{3}\)
2 \(\lambda_{2}=\dfrac{\lambda_{1} \lambda_{3}}{\lambda_{1}+\lambda_{3}}\)
3 \(\lambda_{3}=\dfrac{\lambda_{1} \lambda_{2}}{\lambda_{1}+\lambda_{2}}\)
4 \(\lambda_{1}=\dfrac{\lambda_{2} \lambda_{3}}{\lambda_{2}+\lambda_{3}}\)
KCET - 2023
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PHXII12:ATOMS

356633 Calculate the highest frequency of the emitted photon in the Paschen series of spectral lines of the hydrogen atom.

1 \(3.7 \times {10^{14}}\;Hz\)
2 \(9.1 \times {10^{15}}\;Hz\)
3 \(10.23 \times {10^{14}}\;Hz\)
4 \(29.7 \times {10^{15}}\;Hz\)
PHXII12:ATOMS

356634 In hydrogen spectrum, the shortest wavelength in the Balmer series is \(\lambda\). The shortset wavelength in the Bracket series is:

1 \(4 \lambda\)
2 \(9 \lambda\)
3 \(16 \lambda\)
4 \(2 \lambda\)
NEET - 2023
PHXII12:ATOMS

356635 Hydrogen atom from excited state comes to the ground state by emitting a photon of wavelength \(\lambda \).If \(R\) is the Rydberg constant, the principal quantum number \(n\) of the excited state is

1 \(\sqrt {\frac{{\lambda R}}{{\lambda R - 1}}} \)
2 \(\sqrt {\frac{\lambda }{{\lambda R - 1}}} \)
3 \(\sqrt {\frac{{\lambda {R^2}}}{{\lambda R - 1}}} \)
4 \(\sqrt {\frac{{\lambda R}}{{\lambda - 1}}} \)
KCET - 2009
PHXII12:ATOMS

356636 Three energy levels of hydrogen atom and the corresponding wavelength of the emitted radiation due to different electron transition are as shown. Then.
supporting img

1 \(\lambda_{2}=\lambda_{1}+\lambda_{3}\)
2 \(\lambda_{2}=\dfrac{\lambda_{1} \lambda_{3}}{\lambda_{1}+\lambda_{3}}\)
3 \(\lambda_{3}=\dfrac{\lambda_{1} \lambda_{2}}{\lambda_{1}+\lambda_{2}}\)
4 \(\lambda_{1}=\dfrac{\lambda_{2} \lambda_{3}}{\lambda_{2}+\lambda_{3}}\)
KCET - 2023
NEET DIGITAL LIBRARY
PHXII12:ATOMS

356633 Calculate the highest frequency of the emitted photon in the Paschen series of spectral lines of the hydrogen atom.

1 \(3.7 \times {10^{14}}\;Hz\)
2 \(9.1 \times {10^{15}}\;Hz\)
3 \(10.23 \times {10^{14}}\;Hz\)
4 \(29.7 \times {10^{15}}\;Hz\)
PHXII12:ATOMS

356634 In hydrogen spectrum, the shortest wavelength in the Balmer series is \(\lambda\). The shortset wavelength in the Bracket series is:

1 \(4 \lambda\)
2 \(9 \lambda\)
3 \(16 \lambda\)
4 \(2 \lambda\)
NEET - 2023
PHXII12:ATOMS

356635 Hydrogen atom from excited state comes to the ground state by emitting a photon of wavelength \(\lambda \).If \(R\) is the Rydberg constant, the principal quantum number \(n\) of the excited state is

1 \(\sqrt {\frac{{\lambda R}}{{\lambda R - 1}}} \)
2 \(\sqrt {\frac{\lambda }{{\lambda R - 1}}} \)
3 \(\sqrt {\frac{{\lambda {R^2}}}{{\lambda R - 1}}} \)
4 \(\sqrt {\frac{{\lambda R}}{{\lambda - 1}}} \)
KCET - 2009
PHXII12:ATOMS

356636 Three energy levels of hydrogen atom and the corresponding wavelength of the emitted radiation due to different electron transition are as shown. Then.
supporting img

1 \(\lambda_{2}=\lambda_{1}+\lambda_{3}\)
2 \(\lambda_{2}=\dfrac{\lambda_{1} \lambda_{3}}{\lambda_{1}+\lambda_{3}}\)
3 \(\lambda_{3}=\dfrac{\lambda_{1} \lambda_{2}}{\lambda_{1}+\lambda_{2}}\)
4 \(\lambda_{1}=\dfrac{\lambda_{2} \lambda_{3}}{\lambda_{2}+\lambda_{3}}\)
KCET - 2023
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