The Line Spectra of the Hydrogen Atom
« Previous Topic: Miscellaneous Problems on Atoms Next Topic: X Rays »
PHXII12:ATOMS

356624 The third line of Balmer series of an ion equivalent to hydrogen atom has wavelength of \(108.5\;nm\). The ground state energy of an electron of this ion will be

1 \(3.4\,eV\)
2 \(13.6\,eV\)
3 \(54.4\,eV\)
4 \(122.4\,eV\)
PHXII12:ATOMS

356625 The wavelength of the first line in Balmer series in hydrogen spectrum is \('\lambda '\). What is the wavelength of the second line in the same series?

1 \(\frac{{20}}{{27}}\lambda \)
2 \(\frac{3}{{16}}\lambda \)
3 \(\frac{5}{{16}}\lambda \)
4 \(\frac{3}{4}\lambda \)
MHTCET - 2019
PHXII12:ATOMS

356626 Energy of an electron in the second orbit of hydrogen atom is \({E_2}\) . The energy of electron in the third orbit of \(H{e^ + }\) will be

1 \(\frac{9}{{16}}{E_2}\)
2 \(\frac{{16}}{9}{E_2}\)
3 \(\frac{3}{{16}}{E_2}\)
4 \(\frac{{16}}{3}{E_2}\)
PHXII12:ATOMS

356627 \({v_1}\) is the frequency of the series limit of Lyman series, \({v_2}\) is the frequency of the first line of Lyman series and \({v_3}\) is the frequency of the series limit of the Balmer series. Then

1 \({v_1} - {v_2} = {v_3}\)
2 \({v_1} = {v_2} - {v_3}\)
3 \(\frac{1}{{{v_2}}} = \frac{1}{{{v_1}}} + \frac{1}{{{v_3}}}\)
4 \(\frac{1}{{{v_1}}} = \frac{1}{{{v_2}}} + \frac{1}{{{v_3}}}\)
KCET - 2010
PHXII12:ATOMS

356628 If an electron in hydrogen atom jumps from an orbit of level \(n=3\) to an orbit of level \(n=2\) then emitted radiation has a frequency:

1 \(\dfrac{R c}{9}\)
2 \(\dfrac{R c}{25}\)
3 \(\dfrac{8 R c}{9}\)
4 \(\dfrac{5 R c}{36}\)
Join With Us for More Updates
PHXII12:ATOMS

356624 The third line of Balmer series of an ion equivalent to hydrogen atom has wavelength of \(108.5\;nm\). The ground state energy of an electron of this ion will be

1 \(3.4\,eV\)
2 \(13.6\,eV\)
3 \(54.4\,eV\)
4 \(122.4\,eV\)
PHXII12:ATOMS

356625 The wavelength of the first line in Balmer series in hydrogen spectrum is \('\lambda '\). What is the wavelength of the second line in the same series?

1 \(\frac{{20}}{{27}}\lambda \)
2 \(\frac{3}{{16}}\lambda \)
3 \(\frac{5}{{16}}\lambda \)
4 \(\frac{3}{4}\lambda \)
MHTCET - 2019
PHXII12:ATOMS

356626 Energy of an electron in the second orbit of hydrogen atom is \({E_2}\) . The energy of electron in the third orbit of \(H{e^ + }\) will be

1 \(\frac{9}{{16}}{E_2}\)
2 \(\frac{{16}}{9}{E_2}\)
3 \(\frac{3}{{16}}{E_2}\)
4 \(\frac{{16}}{3}{E_2}\)
PHXII12:ATOMS

356627 \({v_1}\) is the frequency of the series limit of Lyman series, \({v_2}\) is the frequency of the first line of Lyman series and \({v_3}\) is the frequency of the series limit of the Balmer series. Then

1 \({v_1} - {v_2} = {v_3}\)
2 \({v_1} = {v_2} - {v_3}\)
3 \(\frac{1}{{{v_2}}} = \frac{1}{{{v_1}}} + \frac{1}{{{v_3}}}\)
4 \(\frac{1}{{{v_1}}} = \frac{1}{{{v_2}}} + \frac{1}{{{v_3}}}\)
KCET - 2010
PHXII12:ATOMS

356628 If an electron in hydrogen atom jumps from an orbit of level \(n=3\) to an orbit of level \(n=2\) then emitted radiation has a frequency:

1 \(\dfrac{R c}{9}\)
2 \(\dfrac{R c}{25}\)
3 \(\dfrac{8 R c}{9}\)
4 \(\dfrac{5 R c}{36}\)
For More Updates join with Us
NEET Test Series from KOTA - 10 Papers In MS WORD WhatsApp Here
PHXII12:ATOMS

356624 The third line of Balmer series of an ion equivalent to hydrogen atom has wavelength of \(108.5\;nm\). The ground state energy of an electron of this ion will be

1 \(3.4\,eV\)
2 \(13.6\,eV\)
3 \(54.4\,eV\)
4 \(122.4\,eV\)
PHXII12:ATOMS

356625 The wavelength of the first line in Balmer series in hydrogen spectrum is \('\lambda '\). What is the wavelength of the second line in the same series?

1 \(\frac{{20}}{{27}}\lambda \)
2 \(\frac{3}{{16}}\lambda \)
3 \(\frac{5}{{16}}\lambda \)
4 \(\frac{3}{4}\lambda \)
MHTCET - 2019
PHXII12:ATOMS

356626 Energy of an electron in the second orbit of hydrogen atom is \({E_2}\) . The energy of electron in the third orbit of \(H{e^ + }\) will be

1 \(\frac{9}{{16}}{E_2}\)
2 \(\frac{{16}}{9}{E_2}\)
3 \(\frac{3}{{16}}{E_2}\)
4 \(\frac{{16}}{3}{E_2}\)
PHXII12:ATOMS

356627 \({v_1}\) is the frequency of the series limit of Lyman series, \({v_2}\) is the frequency of the first line of Lyman series and \({v_3}\) is the frequency of the series limit of the Balmer series. Then

1 \({v_1} - {v_2} = {v_3}\)
2 \({v_1} = {v_2} - {v_3}\)
3 \(\frac{1}{{{v_2}}} = \frac{1}{{{v_1}}} + \frac{1}{{{v_3}}}\)
4 \(\frac{1}{{{v_1}}} = \frac{1}{{{v_2}}} + \frac{1}{{{v_3}}}\)
KCET - 2010
PHXII12:ATOMS

356628 If an electron in hydrogen atom jumps from an orbit of level \(n=3\) to an orbit of level \(n=2\) then emitted radiation has a frequency:

1 \(\dfrac{R c}{9}\)
2 \(\dfrac{R c}{25}\)
3 \(\dfrac{8 R c}{9}\)
4 \(\dfrac{5 R c}{36}\)
NEET DIGITAL LIBRARY
PHXII12:ATOMS

356624 The third line of Balmer series of an ion equivalent to hydrogen atom has wavelength of \(108.5\;nm\). The ground state energy of an electron of this ion will be

1 \(3.4\,eV\)
2 \(13.6\,eV\)
3 \(54.4\,eV\)
4 \(122.4\,eV\)
PHXII12:ATOMS

356625 The wavelength of the first line in Balmer series in hydrogen spectrum is \('\lambda '\). What is the wavelength of the second line in the same series?

1 \(\frac{{20}}{{27}}\lambda \)
2 \(\frac{3}{{16}}\lambda \)
3 \(\frac{5}{{16}}\lambda \)
4 \(\frac{3}{4}\lambda \)
MHTCET - 2019
PHXII12:ATOMS

356626 Energy of an electron in the second orbit of hydrogen atom is \({E_2}\) . The energy of electron in the third orbit of \(H{e^ + }\) will be

1 \(\frac{9}{{16}}{E_2}\)
2 \(\frac{{16}}{9}{E_2}\)
3 \(\frac{3}{{16}}{E_2}\)
4 \(\frac{{16}}{3}{E_2}\)
PHXII12:ATOMS

356627 \({v_1}\) is the frequency of the series limit of Lyman series, \({v_2}\) is the frequency of the first line of Lyman series and \({v_3}\) is the frequency of the series limit of the Balmer series. Then

1 \({v_1} - {v_2} = {v_3}\)
2 \({v_1} = {v_2} - {v_3}\)
3 \(\frac{1}{{{v_2}}} = \frac{1}{{{v_1}}} + \frac{1}{{{v_3}}}\)
4 \(\frac{1}{{{v_1}}} = \frac{1}{{{v_2}}} + \frac{1}{{{v_3}}}\)
KCET - 2010
PHXII12:ATOMS

356628 If an electron in hydrogen atom jumps from an orbit of level \(n=3\) to an orbit of level \(n=2\) then emitted radiation has a frequency:

1 \(\dfrac{R c}{9}\)
2 \(\dfrac{R c}{25}\)
3 \(\dfrac{8 R c}{9}\)
4 \(\dfrac{5 R c}{36}\)
Join With Us for More Updates
PHXII12:ATOMS

356624 The third line of Balmer series of an ion equivalent to hydrogen atom has wavelength of \(108.5\;nm\). The ground state energy of an electron of this ion will be

1 \(3.4\,eV\)
2 \(13.6\,eV\)
3 \(54.4\,eV\)
4 \(122.4\,eV\)
PHXII12:ATOMS

356625 The wavelength of the first line in Balmer series in hydrogen spectrum is \('\lambda '\). What is the wavelength of the second line in the same series?

1 \(\frac{{20}}{{27}}\lambda \)
2 \(\frac{3}{{16}}\lambda \)
3 \(\frac{5}{{16}}\lambda \)
4 \(\frac{3}{4}\lambda \)
MHTCET - 2019
PHXII12:ATOMS

356626 Energy of an electron in the second orbit of hydrogen atom is \({E_2}\) . The energy of electron in the third orbit of \(H{e^ + }\) will be

1 \(\frac{9}{{16}}{E_2}\)
2 \(\frac{{16}}{9}{E_2}\)
3 \(\frac{3}{{16}}{E_2}\)
4 \(\frac{{16}}{3}{E_2}\)
PHXII12:ATOMS

356627 \({v_1}\) is the frequency of the series limit of Lyman series, \({v_2}\) is the frequency of the first line of Lyman series and \({v_3}\) is the frequency of the series limit of the Balmer series. Then

1 \({v_1} - {v_2} = {v_3}\)
2 \({v_1} = {v_2} - {v_3}\)
3 \(\frac{1}{{{v_2}}} = \frac{1}{{{v_1}}} + \frac{1}{{{v_3}}}\)
4 \(\frac{1}{{{v_1}}} = \frac{1}{{{v_2}}} + \frac{1}{{{v_3}}}\)
KCET - 2010
PHXII12:ATOMS

356628 If an electron in hydrogen atom jumps from an orbit of level \(n=3\) to an orbit of level \(n=2\) then emitted radiation has a frequency:

1 \(\dfrac{R c}{9}\)
2 \(\dfrac{R c}{25}\)
3 \(\dfrac{8 R c}{9}\)
4 \(\dfrac{5 R c}{36}\)