356607
If a hydrogen atom initially at the ground state absorbs a photon and jumps to \(4^{\text {th }}\) excited state then the wavelength of photon is \(9.48 \times 10^{{N}}\). Find the value of ' \(N\) ' is (Take Planck's constant \(=6.6 \times 10^{-34} {Js}\) )
1 \( - 4\)
2 \( - 8\)
3 \( - 12\)
4 \( - 10\)
Explanation:
Here,\({n_1} = 1,\) and \(n_{2}=5\) Energy of absorbed photon, \({E}={E}_{2}-{E}_{1}\) Since, \(E_{n}=-\dfrac{13.6}{n^{2}} e V\) Then, \({E}_{2}-{E}_{1}=-\dfrac{13.6}{(5)^{2}}-\left[-\dfrac{13.6}{(1)^{2}}\right]\) \(=-\dfrac{13.6}{25}+13.6\) \(=-0.544+13.6\) \(=13.056 \,\,{eV}\) \(=13.056 \times 1.6 \times 10^{-19} {~J}\) But \(E_{2}-E_{1}=\dfrac{h c}{\lambda}\) \(\therefore \lambda=\dfrac{h c}{{E}_{2}-{E}_{1}}=\dfrac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{13.056 \times 1.6 \times 10^{-19}}\) \(=9.48 \times 10^{-8} {~m}\) \(\therefore \quad {N}=-8\)
PHXII12:ATOMS
356608
The ratio of the shortest wavelength of Balmer series to the shortest wavelength of Lyman series for hydrogen atom is
1 \(1: 2\)
2 \(4: 1\)
3 \(2: 1\)
4 \(1: 4\)
Explanation:
For transition \(n_{2} \rightarrow n_{1}\), wavelength is givenby, \(\dfrac{1}{\lambda}=R Z^{2}\left(\dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}}\right)\) For shortest wavelength of Lyman series, \(n_{1}=1\) and \(n_{2}=\infty\) \(\therefore \dfrac{1}{\lambda_{L}}=R Z^{2}\left[\dfrac{1}{(1)^{2}}-\dfrac{1}{(\infty)^{2}}\right]=R Z^{2}\) \( \Rightarrow {\lambda _L} = \frac{1}{{R{Z^2}}}\quad \quad \quad (1)\) For shortest wavelength of Balmer series, \(n_{1}=2\) and \(n_{2}=\infty\). \(\therefore \dfrac{1}{\lambda_{B}}=R Z^{2}\left[\dfrac{1}{(2)^{2}}-\dfrac{1}{(\infty)^{2}}\right]=\dfrac{R Z^{2}}{4}\) \( \Rightarrow {\lambda _B} = \frac{4}{{R{Z^2}}}\quad \quad \quad (2)\) Dividing equation (2) by equation (1), we have \(\dfrac{\lambda_{B}}{\lambda_{L}}=\dfrac{4}{R Z^{2}} \times R Z^{2}=\dfrac{4}{1}\)
JEE - 2024
PHXII12:ATOMS
356609
If \(n\) is the orbit number of the electron in a hydrogen atom, the correct statement among the following is
1 Electron energy increases as \(n\) increases
2 Hydrogen emits infrared rays for the electron transition from \(n = \infty \) to \(n = 1\)
3 Electron energy is zero for \(n = 1\)
4 Electron energy varies as \({n^2}\)
Explanation:
In a hydrogen atom the energy of electrons in \({n^{th}}\) orbit is \({E_n} = - \frac{{13.6}}{{{n^2}}}eV\) From the above expression it is clear that electron
KCET - 2012
PHXII12:ATOMS
356610
The first line of Balmer series has wavelength 6563 \( \mathop A^{~~\circ} \). What will be the wavelength of the first member of Lyman series?
1 1215.4 \( \mathop A^{~~\circ} \)
2 2500 \( \mathop A^{~~\circ} \)
3 7500 \( \mathop A^{~~\circ} \)
4 600 \( \mathop A^{~~\circ} \)
Explanation:
When an electron makes a transition from \({m^{\text {th }}}\) orbit to \({n^{\text {th }}}\) orbit, wavelength of the emitted light is given by \({\dfrac{1}{\lambda}=R\left(\dfrac{1}{n^{2}}-\dfrac{1}{m^{2}}\right)}\) Here, \({R=}\) Rydberg constant For first line of Balmer series of hydrogen atom \({[n=2}\) and \({m=3]}\) \(\frac{1}{\lambda } = R\left( {\frac{1}{4} - \frac{1}{9}} \right) = R\left( {\frac{5}{{36}}} \right){\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} (1)\) For first line of Lyman series, \({n=1}\) and \({m=2}\) \(\frac{1}{{{\lambda ^\prime }}} = R\left( {1 - \frac{1}{4}} \right) = R\left( {\frac{3}{4}} \right){\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} (2)\) Dividing eq (1) by (2), we get \({\dfrac{\lambda^{\prime}}{\lambda}=\dfrac{5}{27}}\) \( \Rightarrow {\lambda ^\prime } = \frac{5}{{27}} \times 6563\)\( \mathop A^{~~\circ} \) \( = 1215.4\)\( \mathop A^{~~\circ} \) . So correct option is (1)
356607
If a hydrogen atom initially at the ground state absorbs a photon and jumps to \(4^{\text {th }}\) excited state then the wavelength of photon is \(9.48 \times 10^{{N}}\). Find the value of ' \(N\) ' is (Take Planck's constant \(=6.6 \times 10^{-34} {Js}\) )
1 \( - 4\)
2 \( - 8\)
3 \( - 12\)
4 \( - 10\)
Explanation:
Here,\({n_1} = 1,\) and \(n_{2}=5\) Energy of absorbed photon, \({E}={E}_{2}-{E}_{1}\) Since, \(E_{n}=-\dfrac{13.6}{n^{2}} e V\) Then, \({E}_{2}-{E}_{1}=-\dfrac{13.6}{(5)^{2}}-\left[-\dfrac{13.6}{(1)^{2}}\right]\) \(=-\dfrac{13.6}{25}+13.6\) \(=-0.544+13.6\) \(=13.056 \,\,{eV}\) \(=13.056 \times 1.6 \times 10^{-19} {~J}\) But \(E_{2}-E_{1}=\dfrac{h c}{\lambda}\) \(\therefore \lambda=\dfrac{h c}{{E}_{2}-{E}_{1}}=\dfrac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{13.056 \times 1.6 \times 10^{-19}}\) \(=9.48 \times 10^{-8} {~m}\) \(\therefore \quad {N}=-8\)
PHXII12:ATOMS
356608
The ratio of the shortest wavelength of Balmer series to the shortest wavelength of Lyman series for hydrogen atom is
1 \(1: 2\)
2 \(4: 1\)
3 \(2: 1\)
4 \(1: 4\)
Explanation:
For transition \(n_{2} \rightarrow n_{1}\), wavelength is givenby, \(\dfrac{1}{\lambda}=R Z^{2}\left(\dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}}\right)\) For shortest wavelength of Lyman series, \(n_{1}=1\) and \(n_{2}=\infty\) \(\therefore \dfrac{1}{\lambda_{L}}=R Z^{2}\left[\dfrac{1}{(1)^{2}}-\dfrac{1}{(\infty)^{2}}\right]=R Z^{2}\) \( \Rightarrow {\lambda _L} = \frac{1}{{R{Z^2}}}\quad \quad \quad (1)\) For shortest wavelength of Balmer series, \(n_{1}=2\) and \(n_{2}=\infty\). \(\therefore \dfrac{1}{\lambda_{B}}=R Z^{2}\left[\dfrac{1}{(2)^{2}}-\dfrac{1}{(\infty)^{2}}\right]=\dfrac{R Z^{2}}{4}\) \( \Rightarrow {\lambda _B} = \frac{4}{{R{Z^2}}}\quad \quad \quad (2)\) Dividing equation (2) by equation (1), we have \(\dfrac{\lambda_{B}}{\lambda_{L}}=\dfrac{4}{R Z^{2}} \times R Z^{2}=\dfrac{4}{1}\)
JEE - 2024
PHXII12:ATOMS
356609
If \(n\) is the orbit number of the electron in a hydrogen atom, the correct statement among the following is
1 Electron energy increases as \(n\) increases
2 Hydrogen emits infrared rays for the electron transition from \(n = \infty \) to \(n = 1\)
3 Electron energy is zero for \(n = 1\)
4 Electron energy varies as \({n^2}\)
Explanation:
In a hydrogen atom the energy of electrons in \({n^{th}}\) orbit is \({E_n} = - \frac{{13.6}}{{{n^2}}}eV\) From the above expression it is clear that electron
KCET - 2012
PHXII12:ATOMS
356610
The first line of Balmer series has wavelength 6563 \( \mathop A^{~~\circ} \). What will be the wavelength of the first member of Lyman series?
1 1215.4 \( \mathop A^{~~\circ} \)
2 2500 \( \mathop A^{~~\circ} \)
3 7500 \( \mathop A^{~~\circ} \)
4 600 \( \mathop A^{~~\circ} \)
Explanation:
When an electron makes a transition from \({m^{\text {th }}}\) orbit to \({n^{\text {th }}}\) orbit, wavelength of the emitted light is given by \({\dfrac{1}{\lambda}=R\left(\dfrac{1}{n^{2}}-\dfrac{1}{m^{2}}\right)}\) Here, \({R=}\) Rydberg constant For first line of Balmer series of hydrogen atom \({[n=2}\) and \({m=3]}\) \(\frac{1}{\lambda } = R\left( {\frac{1}{4} - \frac{1}{9}} \right) = R\left( {\frac{5}{{36}}} \right){\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} (1)\) For first line of Lyman series, \({n=1}\) and \({m=2}\) \(\frac{1}{{{\lambda ^\prime }}} = R\left( {1 - \frac{1}{4}} \right) = R\left( {\frac{3}{4}} \right){\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} (2)\) Dividing eq (1) by (2), we get \({\dfrac{\lambda^{\prime}}{\lambda}=\dfrac{5}{27}}\) \( \Rightarrow {\lambda ^\prime } = \frac{5}{{27}} \times 6563\)\( \mathop A^{~~\circ} \) \( = 1215.4\)\( \mathop A^{~~\circ} \) . So correct option is (1)
356607
If a hydrogen atom initially at the ground state absorbs a photon and jumps to \(4^{\text {th }}\) excited state then the wavelength of photon is \(9.48 \times 10^{{N}}\). Find the value of ' \(N\) ' is (Take Planck's constant \(=6.6 \times 10^{-34} {Js}\) )
1 \( - 4\)
2 \( - 8\)
3 \( - 12\)
4 \( - 10\)
Explanation:
Here,\({n_1} = 1,\) and \(n_{2}=5\) Energy of absorbed photon, \({E}={E}_{2}-{E}_{1}\) Since, \(E_{n}=-\dfrac{13.6}{n^{2}} e V\) Then, \({E}_{2}-{E}_{1}=-\dfrac{13.6}{(5)^{2}}-\left[-\dfrac{13.6}{(1)^{2}}\right]\) \(=-\dfrac{13.6}{25}+13.6\) \(=-0.544+13.6\) \(=13.056 \,\,{eV}\) \(=13.056 \times 1.6 \times 10^{-19} {~J}\) But \(E_{2}-E_{1}=\dfrac{h c}{\lambda}\) \(\therefore \lambda=\dfrac{h c}{{E}_{2}-{E}_{1}}=\dfrac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{13.056 \times 1.6 \times 10^{-19}}\) \(=9.48 \times 10^{-8} {~m}\) \(\therefore \quad {N}=-8\)
PHXII12:ATOMS
356608
The ratio of the shortest wavelength of Balmer series to the shortest wavelength of Lyman series for hydrogen atom is
1 \(1: 2\)
2 \(4: 1\)
3 \(2: 1\)
4 \(1: 4\)
Explanation:
For transition \(n_{2} \rightarrow n_{1}\), wavelength is givenby, \(\dfrac{1}{\lambda}=R Z^{2}\left(\dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}}\right)\) For shortest wavelength of Lyman series, \(n_{1}=1\) and \(n_{2}=\infty\) \(\therefore \dfrac{1}{\lambda_{L}}=R Z^{2}\left[\dfrac{1}{(1)^{2}}-\dfrac{1}{(\infty)^{2}}\right]=R Z^{2}\) \( \Rightarrow {\lambda _L} = \frac{1}{{R{Z^2}}}\quad \quad \quad (1)\) For shortest wavelength of Balmer series, \(n_{1}=2\) and \(n_{2}=\infty\). \(\therefore \dfrac{1}{\lambda_{B}}=R Z^{2}\left[\dfrac{1}{(2)^{2}}-\dfrac{1}{(\infty)^{2}}\right]=\dfrac{R Z^{2}}{4}\) \( \Rightarrow {\lambda _B} = \frac{4}{{R{Z^2}}}\quad \quad \quad (2)\) Dividing equation (2) by equation (1), we have \(\dfrac{\lambda_{B}}{\lambda_{L}}=\dfrac{4}{R Z^{2}} \times R Z^{2}=\dfrac{4}{1}\)
JEE - 2024
PHXII12:ATOMS
356609
If \(n\) is the orbit number of the electron in a hydrogen atom, the correct statement among the following is
1 Electron energy increases as \(n\) increases
2 Hydrogen emits infrared rays for the electron transition from \(n = \infty \) to \(n = 1\)
3 Electron energy is zero for \(n = 1\)
4 Electron energy varies as \({n^2}\)
Explanation:
In a hydrogen atom the energy of electrons in \({n^{th}}\) orbit is \({E_n} = - \frac{{13.6}}{{{n^2}}}eV\) From the above expression it is clear that electron
KCET - 2012
PHXII12:ATOMS
356610
The first line of Balmer series has wavelength 6563 \( \mathop A^{~~\circ} \). What will be the wavelength of the first member of Lyman series?
1 1215.4 \( \mathop A^{~~\circ} \)
2 2500 \( \mathop A^{~~\circ} \)
3 7500 \( \mathop A^{~~\circ} \)
4 600 \( \mathop A^{~~\circ} \)
Explanation:
When an electron makes a transition from \({m^{\text {th }}}\) orbit to \({n^{\text {th }}}\) orbit, wavelength of the emitted light is given by \({\dfrac{1}{\lambda}=R\left(\dfrac{1}{n^{2}}-\dfrac{1}{m^{2}}\right)}\) Here, \({R=}\) Rydberg constant For first line of Balmer series of hydrogen atom \({[n=2}\) and \({m=3]}\) \(\frac{1}{\lambda } = R\left( {\frac{1}{4} - \frac{1}{9}} \right) = R\left( {\frac{5}{{36}}} \right){\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} (1)\) For first line of Lyman series, \({n=1}\) and \({m=2}\) \(\frac{1}{{{\lambda ^\prime }}} = R\left( {1 - \frac{1}{4}} \right) = R\left( {\frac{3}{4}} \right){\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} (2)\) Dividing eq (1) by (2), we get \({\dfrac{\lambda^{\prime}}{\lambda}=\dfrac{5}{27}}\) \( \Rightarrow {\lambda ^\prime } = \frac{5}{{27}} \times 6563\)\( \mathop A^{~~\circ} \) \( = 1215.4\)\( \mathop A^{~~\circ} \) . So correct option is (1)
356607
If a hydrogen atom initially at the ground state absorbs a photon and jumps to \(4^{\text {th }}\) excited state then the wavelength of photon is \(9.48 \times 10^{{N}}\). Find the value of ' \(N\) ' is (Take Planck's constant \(=6.6 \times 10^{-34} {Js}\) )
1 \( - 4\)
2 \( - 8\)
3 \( - 12\)
4 \( - 10\)
Explanation:
Here,\({n_1} = 1,\) and \(n_{2}=5\) Energy of absorbed photon, \({E}={E}_{2}-{E}_{1}\) Since, \(E_{n}=-\dfrac{13.6}{n^{2}} e V\) Then, \({E}_{2}-{E}_{1}=-\dfrac{13.6}{(5)^{2}}-\left[-\dfrac{13.6}{(1)^{2}}\right]\) \(=-\dfrac{13.6}{25}+13.6\) \(=-0.544+13.6\) \(=13.056 \,\,{eV}\) \(=13.056 \times 1.6 \times 10^{-19} {~J}\) But \(E_{2}-E_{1}=\dfrac{h c}{\lambda}\) \(\therefore \lambda=\dfrac{h c}{{E}_{2}-{E}_{1}}=\dfrac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{13.056 \times 1.6 \times 10^{-19}}\) \(=9.48 \times 10^{-8} {~m}\) \(\therefore \quad {N}=-8\)
PHXII12:ATOMS
356608
The ratio of the shortest wavelength of Balmer series to the shortest wavelength of Lyman series for hydrogen atom is
1 \(1: 2\)
2 \(4: 1\)
3 \(2: 1\)
4 \(1: 4\)
Explanation:
For transition \(n_{2} \rightarrow n_{1}\), wavelength is givenby, \(\dfrac{1}{\lambda}=R Z^{2}\left(\dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}}\right)\) For shortest wavelength of Lyman series, \(n_{1}=1\) and \(n_{2}=\infty\) \(\therefore \dfrac{1}{\lambda_{L}}=R Z^{2}\left[\dfrac{1}{(1)^{2}}-\dfrac{1}{(\infty)^{2}}\right]=R Z^{2}\) \( \Rightarrow {\lambda _L} = \frac{1}{{R{Z^2}}}\quad \quad \quad (1)\) For shortest wavelength of Balmer series, \(n_{1}=2\) and \(n_{2}=\infty\). \(\therefore \dfrac{1}{\lambda_{B}}=R Z^{2}\left[\dfrac{1}{(2)^{2}}-\dfrac{1}{(\infty)^{2}}\right]=\dfrac{R Z^{2}}{4}\) \( \Rightarrow {\lambda _B} = \frac{4}{{R{Z^2}}}\quad \quad \quad (2)\) Dividing equation (2) by equation (1), we have \(\dfrac{\lambda_{B}}{\lambda_{L}}=\dfrac{4}{R Z^{2}} \times R Z^{2}=\dfrac{4}{1}\)
JEE - 2024
PHXII12:ATOMS
356609
If \(n\) is the orbit number of the electron in a hydrogen atom, the correct statement among the following is
1 Electron energy increases as \(n\) increases
2 Hydrogen emits infrared rays for the electron transition from \(n = \infty \) to \(n = 1\)
3 Electron energy is zero for \(n = 1\)
4 Electron energy varies as \({n^2}\)
Explanation:
In a hydrogen atom the energy of electrons in \({n^{th}}\) orbit is \({E_n} = - \frac{{13.6}}{{{n^2}}}eV\) From the above expression it is clear that electron
KCET - 2012
PHXII12:ATOMS
356610
The first line of Balmer series has wavelength 6563 \( \mathop A^{~~\circ} \). What will be the wavelength of the first member of Lyman series?
1 1215.4 \( \mathop A^{~~\circ} \)
2 2500 \( \mathop A^{~~\circ} \)
3 7500 \( \mathop A^{~~\circ} \)
4 600 \( \mathop A^{~~\circ} \)
Explanation:
When an electron makes a transition from \({m^{\text {th }}}\) orbit to \({n^{\text {th }}}\) orbit, wavelength of the emitted light is given by \({\dfrac{1}{\lambda}=R\left(\dfrac{1}{n^{2}}-\dfrac{1}{m^{2}}\right)}\) Here, \({R=}\) Rydberg constant For first line of Balmer series of hydrogen atom \({[n=2}\) and \({m=3]}\) \(\frac{1}{\lambda } = R\left( {\frac{1}{4} - \frac{1}{9}} \right) = R\left( {\frac{5}{{36}}} \right){\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} (1)\) For first line of Lyman series, \({n=1}\) and \({m=2}\) \(\frac{1}{{{\lambda ^\prime }}} = R\left( {1 - \frac{1}{4}} \right) = R\left( {\frac{3}{4}} \right){\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} (2)\) Dividing eq (1) by (2), we get \({\dfrac{\lambda^{\prime}}{\lambda}=\dfrac{5}{27}}\) \( \Rightarrow {\lambda ^\prime } = \frac{5}{{27}} \times 6563\)\( \mathop A^{~~\circ} \) \( = 1215.4\)\( \mathop A^{~~\circ} \) . So correct option is (1)
