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PHXII12:ATOMS

356551 Electrons in a certain energy level $n = n_{1}$ , can emit 3 spectral lines. When they are in another energy level $n = n_{2}$ . They can emit 6 spectral lines. The orbital speed of the electrons in the two orbits are in the ratio

1

4 : 3

2

3 : 4

3

2 : 1

4

1 : 2

Explanation:

Number of emission spectral lines
$N = \frac{n (n - 1)}{2}$
$∴ 3 = \frac{n_{1} (n_{1} - 1)}{2}$ in first case.
or $n_{1}^{2} - n_{1} - 6 = 0 \Rightarrow (n_{1} - 3) (n_{1} + 2) = 0$
Taking positive root
$∴ n_{1} = 3$ (1)
In second case, $6 = \frac{n_{2} (n_{2} - 1)}{2}$
or $n_{2}^{2} - n_{2} - 12 = 0 \Rightarrow (n_{2} - 4) (n_{2} + 3) = 0$
Taking positive root, $n_{2} = 4$ (2)
Velocity of electron, $v = \frac{2 π K Z e^{2}}{n h}$
$∴ \frac{v_{1}}{v_{2}} = \frac{n_{2}}{n_{1}} = \frac{4}{3}$

KCET - 2007

PHXII12:ATOMS

356552 Figure represents in simplified form of some lower energy levels of the hydrogen atom. If the transition of an electron from $E_{1} t o E_{2}$ were associated with the emision of blue light, which one of the following transitions could be associated with the emission of red light?
supporting img

1

E_{4} t o E_{1}

2

E_{3} t o E_{1}

3

E_{3} t o E_{2}

4

E_{1} t o E_{3}

Explanation:

Blue light falls in the Balmer series and it is obtained when the atom makes a transition from $E_{4} t o E_{3}$ . Red light also falls in the Balmer series and it has a lower frequency compared to blue light. By quantum theory of radiation, $E = h f$ . Thus, red light is associated with a smaller energy change from a lower energy level (compared to $E_{4}$ ) to the first excited state $E_{2}$ . Hence, the only possible transition that results in the emission of red light is the $E_{3} t o E_{2}$ transition.

PHXII12:ATOMS

356553 The ratio of energies of photons produced due to transition of electron of hydrogen atom from its (i) second to first energy level and (ii) highest energy level to second level is respectively

1

2.5 : 1

2

3 : 1

3

2 : 1

4

4 : 1

Explanation:

The energy of photons is given by
$E = R h c (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$
Where, $R$ is Rydberg constant, $h$ is Planck’s constant and $c$ is the speed of light
Energy of photon produced from second to first energy level
$E_{1} = R h c (\frac{1}{1^{2}} - \frac{1}{2^{2}})$
$E_{1} = \frac{3}{4} R h c$
Energy of photon produced from higher energy level (i.e., $\infty$ ) to second level
$E_{2} = R h c (\frac{1}{2^{2}} - \frac{1}{\infty}) = \frac{1}{4} R h c$
$∴ \frac{E_{1}}{E_{2}} = \frac{\frac{3}{4} R h c}{\frac{1}{4} R h c} = \frac{3}{1} o r 3 : 1$

MHTCET - 2020

PHXII12:ATOMS

356554 The transition from the state $n = 3$ to $n = 1$ in a hydrogen like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition from

1

4 \to 2

2

4 \to 3

3

2 \to 1

4

3 \to 2

Explanation:

Lyman series produces $U V$ radiation. Balmer series produces visible radiation. Pachen series produces infrared radiation Least energetic photon in Paschen series is obtained from $n_{i} = 4$ to $n_{f} = 3$

PHXII12:ATOMS

356555 If wavelength of photon emitted due to transition of an electron from the third orbit to the first orbit in a hydrogen atom is $λ$ , then the wavelength of photon emitted due to transition of electrons from the fourth orbit to the second orbit will be $N$ times $λ$ . What is the value of $N$ ?

1

2.65 λ

2

7.35 λ

3

5.47 λ

4

4.74 λ

Explanation:

Transition of electrons from the third orbit to the first orbit,
$\frac{1}{λ} = R [\frac{1}{1^{2}} - \frac{1}{3^{2}}]$
$\Rightarrow λ = \frac{9}{8 R}$
Transition of electrons from the fourth orbit to the second orbit,
$\frac{1}{λ^{'}} = R [\frac{1}{2^{2}} - \frac{1}{4^{2}}] \Rightarrow λ^{'} = \frac{16}{3 R}$
Now, $\frac{λ^{'}}{λ} = \frac{16}{3 R} \times \frac{8 R}{9} = \frac{128}{27}$
$\Rightarrow λ^{'} = \frac{128}{27} λ = 4.74 λ$

PHXII12:ATOMS

356551 Electrons in a certain energy level $n = n_{1}$ , can emit 3 spectral lines. When they are in another energy level $n = n_{2}$ . They can emit 6 spectral lines. The orbital speed of the electrons in the two orbits are in the ratio

1

4 : 3

2

3 : 4

3

2 : 1

4

1 : 2

Explanation:

Number of emission spectral lines
$N = \frac{n (n - 1)}{2}$
$∴ 3 = \frac{n_{1} (n_{1} - 1)}{2}$ in first case.
or $n_{1}^{2} - n_{1} - 6 = 0 \Rightarrow (n_{1} - 3) (n_{1} + 2) = 0$
Taking positive root
$∴ n_{1} = 3$ (1)
In second case, $6 = \frac{n_{2} (n_{2} - 1)}{2}$
or $n_{2}^{2} - n_{2} - 12 = 0 \Rightarrow (n_{2} - 4) (n_{2} + 3) = 0$
Taking positive root, $n_{2} = 4$ (2)
Velocity of electron, $v = \frac{2 π K Z e^{2}}{n h}$
$∴ \frac{v_{1}}{v_{2}} = \frac{n_{2}}{n_{1}} = \frac{4}{3}$

KCET - 2007

PHXII12:ATOMS

356552 Figure represents in simplified form of some lower energy levels of the hydrogen atom. If the transition of an electron from $E_{1} t o E_{2}$ were associated with the emision of blue light, which one of the following transitions could be associated with the emission of red light?
supporting img

1

E_{4} t o E_{1}

2

E_{3} t o E_{1}

3

E_{3} t o E_{2}

4

E_{1} t o E_{3}

Explanation:

Blue light falls in the Balmer series and it is obtained when the atom makes a transition from $E_{4} t o E_{3}$ . Red light also falls in the Balmer series and it has a lower frequency compared to blue light. By quantum theory of radiation, $E = h f$ . Thus, red light is associated with a smaller energy change from a lower energy level (compared to $E_{4}$ ) to the first excited state $E_{2}$ . Hence, the only possible transition that results in the emission of red light is the $E_{3} t o E_{2}$ transition.

PHXII12:ATOMS

356553 The ratio of energies of photons produced due to transition of electron of hydrogen atom from its (i) second to first energy level and (ii) highest energy level to second level is respectively

1

2.5 : 1

2

3 : 1

3

2 : 1

4

4 : 1

Explanation:

The energy of photons is given by
$E = R h c (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$
Where, $R$ is Rydberg constant, $h$ is Planck’s constant and $c$ is the speed of light
Energy of photon produced from second to first energy level
$E_{1} = R h c (\frac{1}{1^{2}} - \frac{1}{2^{2}})$
$E_{1} = \frac{3}{4} R h c$
Energy of photon produced from higher energy level (i.e., $\infty$ ) to second level
$E_{2} = R h c (\frac{1}{2^{2}} - \frac{1}{\infty}) = \frac{1}{4} R h c$
$∴ \frac{E_{1}}{E_{2}} = \frac{\frac{3}{4} R h c}{\frac{1}{4} R h c} = \frac{3}{1} o r 3 : 1$

MHTCET - 2020

PHXII12:ATOMS

356554 The transition from the state $n = 3$ to $n = 1$ in a hydrogen like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition from

1

4 \to 2

2

4 \to 3

3

2 \to 1

4

3 \to 2

Explanation:

Lyman series produces $U V$ radiation. Balmer series produces visible radiation. Pachen series produces infrared radiation Least energetic photon in Paschen series is obtained from $n_{i} = 4$ to $n_{f} = 3$

PHXII12:ATOMS

356555 If wavelength of photon emitted due to transition of an electron from the third orbit to the first orbit in a hydrogen atom is $λ$ , then the wavelength of photon emitted due to transition of electrons from the fourth orbit to the second orbit will be $N$ times $λ$ . What is the value of $N$ ?

1

2.65 λ

2

7.35 λ

3

5.47 λ

4

4.74 λ

Explanation:

Transition of electrons from the third orbit to the first orbit,
$\frac{1}{λ} = R [\frac{1}{1^{2}} - \frac{1}{3^{2}}]$
$\Rightarrow λ = \frac{9}{8 R}$
Transition of electrons from the fourth orbit to the second orbit,
$\frac{1}{λ^{'}} = R [\frac{1}{2^{2}} - \frac{1}{4^{2}}] \Rightarrow λ^{'} = \frac{16}{3 R}$
Now, $\frac{λ^{'}}{λ} = \frac{16}{3 R} \times \frac{8 R}{9} = \frac{128}{27}$
$\Rightarrow λ^{'} = \frac{128}{27} λ = 4.74 λ$

PHXII12:ATOMS

356551 Electrons in a certain energy level $n = n_{1}$ , can emit 3 spectral lines. When they are in another energy level $n = n_{2}$ . They can emit 6 spectral lines. The orbital speed of the electrons in the two orbits are in the ratio

1

4 : 3

2

3 : 4

3

2 : 1

4

1 : 2

Explanation:

Number of emission spectral lines
$N = \frac{n (n - 1)}{2}$
$∴ 3 = \frac{n_{1} (n_{1} - 1)}{2}$ in first case.
or $n_{1}^{2} - n_{1} - 6 = 0 \Rightarrow (n_{1} - 3) (n_{1} + 2) = 0$
Taking positive root
$∴ n_{1} = 3$ (1)
In second case, $6 = \frac{n_{2} (n_{2} - 1)}{2}$
or $n_{2}^{2} - n_{2} - 12 = 0 \Rightarrow (n_{2} - 4) (n_{2} + 3) = 0$
Taking positive root, $n_{2} = 4$ (2)
Velocity of electron, $v = \frac{2 π K Z e^{2}}{n h}$
$∴ \frac{v_{1}}{v_{2}} = \frac{n_{2}}{n_{1}} = \frac{4}{3}$

KCET - 2007

PHXII12:ATOMS

356552 Figure represents in simplified form of some lower energy levels of the hydrogen atom. If the transition of an electron from $E_{1} t o E_{2}$ were associated with the emision of blue light, which one of the following transitions could be associated with the emission of red light?
supporting img

1

E_{4} t o E_{1}

2

E_{3} t o E_{1}

3

E_{3} t o E_{2}

4

E_{1} t o E_{3}

Explanation:

Blue light falls in the Balmer series and it is obtained when the atom makes a transition from $E_{4} t o E_{3}$ . Red light also falls in the Balmer series and it has a lower frequency compared to blue light. By quantum theory of radiation, $E = h f$ . Thus, red light is associated with a smaller energy change from a lower energy level (compared to $E_{4}$ ) to the first excited state $E_{2}$ . Hence, the only possible transition that results in the emission of red light is the $E_{3} t o E_{2}$ transition.

PHXII12:ATOMS

356553 The ratio of energies of photons produced due to transition of electron of hydrogen atom from its (i) second to first energy level and (ii) highest energy level to second level is respectively

1

2.5 : 1

2

3 : 1

3

2 : 1

4

4 : 1

Explanation:

The energy of photons is given by
$E = R h c (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$
Where, $R$ is Rydberg constant, $h$ is Planck’s constant and $c$ is the speed of light
Energy of photon produced from second to first energy level
$E_{1} = R h c (\frac{1}{1^{2}} - \frac{1}{2^{2}})$
$E_{1} = \frac{3}{4} R h c$
Energy of photon produced from higher energy level (i.e., $\infty$ ) to second level
$E_{2} = R h c (\frac{1}{2^{2}} - \frac{1}{\infty}) = \frac{1}{4} R h c$
$∴ \frac{E_{1}}{E_{2}} = \frac{\frac{3}{4} R h c}{\frac{1}{4} R h c} = \frac{3}{1} o r 3 : 1$

MHTCET - 2020

PHXII12:ATOMS

356554 The transition from the state $n = 3$ to $n = 1$ in a hydrogen like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition from

1

4 \to 2

2

4 \to 3

3

2 \to 1

4

3 \to 2

Explanation:

Lyman series produces $U V$ radiation. Balmer series produces visible radiation. Pachen series produces infrared radiation Least energetic photon in Paschen series is obtained from $n_{i} = 4$ to $n_{f} = 3$

PHXII12:ATOMS

356555 If wavelength of photon emitted due to transition of an electron from the third orbit to the first orbit in a hydrogen atom is $λ$ , then the wavelength of photon emitted due to transition of electrons from the fourth orbit to the second orbit will be $N$ times $λ$ . What is the value of $N$ ?

1

2.65 λ

2

7.35 λ

3

5.47 λ

4

4.74 λ

Explanation:

Transition of electrons from the third orbit to the first orbit,
$\frac{1}{λ} = R [\frac{1}{1^{2}} - \frac{1}{3^{2}}]$
$\Rightarrow λ = \frac{9}{8 R}$
Transition of electrons from the fourth orbit to the second orbit,
$\frac{1}{λ^{'}} = R [\frac{1}{2^{2}} - \frac{1}{4^{2}}] \Rightarrow λ^{'} = \frac{16}{3 R}$
Now, $\frac{λ^{'}}{λ} = \frac{16}{3 R} \times \frac{8 R}{9} = \frac{128}{27}$
$\Rightarrow λ^{'} = \frac{128}{27} λ = 4.74 λ$

PHXII12:ATOMS

356551 Electrons in a certain energy level $n = n_{1}$ , can emit 3 spectral lines. When they are in another energy level $n = n_{2}$ . They can emit 6 spectral lines. The orbital speed of the electrons in the two orbits are in the ratio

1

4 : 3

2

3 : 4

3

2 : 1

4

1 : 2

Explanation:

Number of emission spectral lines
$N = \frac{n (n - 1)}{2}$
$∴ 3 = \frac{n_{1} (n_{1} - 1)}{2}$ in first case.
or $n_{1}^{2} - n_{1} - 6 = 0 \Rightarrow (n_{1} - 3) (n_{1} + 2) = 0$
Taking positive root
$∴ n_{1} = 3$ (1)
In second case, $6 = \frac{n_{2} (n_{2} - 1)}{2}$
or $n_{2}^{2} - n_{2} - 12 = 0 \Rightarrow (n_{2} - 4) (n_{2} + 3) = 0$
Taking positive root, $n_{2} = 4$ (2)
Velocity of electron, $v = \frac{2 π K Z e^{2}}{n h}$
$∴ \frac{v_{1}}{v_{2}} = \frac{n_{2}}{n_{1}} = \frac{4}{3}$

KCET - 2007

PHXII12:ATOMS

356552 Figure represents in simplified form of some lower energy levels of the hydrogen atom. If the transition of an electron from $E_{1} t o E_{2}$ were associated with the emision of blue light, which one of the following transitions could be associated with the emission of red light?
supporting img

1

E_{4} t o E_{1}

2

E_{3} t o E_{1}

3

E_{3} t o E_{2}

4

E_{1} t o E_{3}

Explanation:

Blue light falls in the Balmer series and it is obtained when the atom makes a transition from $E_{4} t o E_{3}$ . Red light also falls in the Balmer series and it has a lower frequency compared to blue light. By quantum theory of radiation, $E = h f$ . Thus, red light is associated with a smaller energy change from a lower energy level (compared to $E_{4}$ ) to the first excited state $E_{2}$ . Hence, the only possible transition that results in the emission of red light is the $E_{3} t o E_{2}$ transition.

PHXII12:ATOMS

356553 The ratio of energies of photons produced due to transition of electron of hydrogen atom from its (i) second to first energy level and (ii) highest energy level to second level is respectively

1

2.5 : 1

2

3 : 1

3

2 : 1

4

4 : 1

Explanation:

The energy of photons is given by
$E = R h c (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$
Where, $R$ is Rydberg constant, $h$ is Planck’s constant and $c$ is the speed of light
Energy of photon produced from second to first energy level
$E_{1} = R h c (\frac{1}{1^{2}} - \frac{1}{2^{2}})$
$E_{1} = \frac{3}{4} R h c$
Energy of photon produced from higher energy level (i.e., $\infty$ ) to second level
$E_{2} = R h c (\frac{1}{2^{2}} - \frac{1}{\infty}) = \frac{1}{4} R h c$
$∴ \frac{E_{1}}{E_{2}} = \frac{\frac{3}{4} R h c}{\frac{1}{4} R h c} = \frac{3}{1} o r 3 : 1$

MHTCET - 2020

PHXII12:ATOMS

356554 The transition from the state $n = 3$ to $n = 1$ in a hydrogen like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition from

1

4 \to 2

2

4 \to 3

3

2 \to 1

4

3 \to 2

Explanation:

Lyman series produces $U V$ radiation. Balmer series produces visible radiation. Pachen series produces infrared radiation Least energetic photon in Paschen series is obtained from $n_{i} = 4$ to $n_{f} = 3$

PHXII12:ATOMS

356555 If wavelength of photon emitted due to transition of an electron from the third orbit to the first orbit in a hydrogen atom is $λ$ , then the wavelength of photon emitted due to transition of electrons from the fourth orbit to the second orbit will be $N$ times $λ$ . What is the value of $N$ ?

1

2.65 λ

2

7.35 λ

3

5.47 λ

4

4.74 λ

Explanation:

Transition of electrons from the third orbit to the first orbit,
$\frac{1}{λ} = R [\frac{1}{1^{2}} - \frac{1}{3^{2}}]$
$\Rightarrow λ = \frac{9}{8 R}$
Transition of electrons from the fourth orbit to the second orbit,
$\frac{1}{λ^{'}} = R [\frac{1}{2^{2}} - \frac{1}{4^{2}}] \Rightarrow λ^{'} = \frac{16}{3 R}$
Now, $\frac{λ^{'}}{λ} = \frac{16}{3 R} \times \frac{8 R}{9} = \frac{128}{27}$
$\Rightarrow λ^{'} = \frac{128}{27} λ = 4.74 λ$

PHXII12:ATOMS

356551 Electrons in a certain energy level $n = n_{1}$ , can emit 3 spectral lines. When they are in another energy level $n = n_{2}$ . They can emit 6 spectral lines. The orbital speed of the electrons in the two orbits are in the ratio

1

4 : 3

2

3 : 4

3

2 : 1

4

1 : 2

Explanation:

Number of emission spectral lines
$N = \frac{n (n - 1)}{2}$
$∴ 3 = \frac{n_{1} (n_{1} - 1)}{2}$ in first case.
or $n_{1}^{2} - n_{1} - 6 = 0 \Rightarrow (n_{1} - 3) (n_{1} + 2) = 0$
Taking positive root
$∴ n_{1} = 3$ (1)
In second case, $6 = \frac{n_{2} (n_{2} - 1)}{2}$
or $n_{2}^{2} - n_{2} - 12 = 0 \Rightarrow (n_{2} - 4) (n_{2} + 3) = 0$
Taking positive root, $n_{2} = 4$ (2)
Velocity of electron, $v = \frac{2 π K Z e^{2}}{n h}$
$∴ \frac{v_{1}}{v_{2}} = \frac{n_{2}}{n_{1}} = \frac{4}{3}$

KCET - 2007

PHXII12:ATOMS

356552 Figure represents in simplified form of some lower energy levels of the hydrogen atom. If the transition of an electron from $E_{1} t o E_{2}$ were associated with the emision of blue light, which one of the following transitions could be associated with the emission of red light?
supporting img

1

E_{4} t o E_{1}

2

E_{3} t o E_{1}

3

E_{3} t o E_{2}

4

E_{1} t o E_{3}

Explanation:

Blue light falls in the Balmer series and it is obtained when the atom makes a transition from $E_{4} t o E_{3}$ . Red light also falls in the Balmer series and it has a lower frequency compared to blue light. By quantum theory of radiation, $E = h f$ . Thus, red light is associated with a smaller energy change from a lower energy level (compared to $E_{4}$ ) to the first excited state $E_{2}$ . Hence, the only possible transition that results in the emission of red light is the $E_{3} t o E_{2}$ transition.

PHXII12:ATOMS

356553 The ratio of energies of photons produced due to transition of electron of hydrogen atom from its (i) second to first energy level and (ii) highest energy level to second level is respectively

1

2.5 : 1

2

3 : 1

3

2 : 1

4

4 : 1

Explanation:

The energy of photons is given by
$E = R h c (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$
Where, $R$ is Rydberg constant, $h$ is Planck’s constant and $c$ is the speed of light
Energy of photon produced from second to first energy level
$E_{1} = R h c (\frac{1}{1^{2}} - \frac{1}{2^{2}})$
$E_{1} = \frac{3}{4} R h c$
Energy of photon produced from higher energy level (i.e., $\infty$ ) to second level
$E_{2} = R h c (\frac{1}{2^{2}} - \frac{1}{\infty}) = \frac{1}{4} R h c$
$∴ \frac{E_{1}}{E_{2}} = \frac{\frac{3}{4} R h c}{\frac{1}{4} R h c} = \frac{3}{1} o r 3 : 1$

MHTCET - 2020

PHXII12:ATOMS

356554 The transition from the state $n = 3$ to $n = 1$ in a hydrogen like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition from

1

4 \to 2

2

4 \to 3

3

2 \to 1

4

3 \to 2

Explanation:

Lyman series produces $U V$ radiation. Balmer series produces visible radiation. Pachen series produces infrared radiation Least energetic photon in Paschen series is obtained from $n_{i} = 4$ to $n_{f} = 3$

PHXII12:ATOMS

356555 If wavelength of photon emitted due to transition of an electron from the third orbit to the first orbit in a hydrogen atom is $λ$ , then the wavelength of photon emitted due to transition of electrons from the fourth orbit to the second orbit will be $N$ times $λ$ . What is the value of $N$ ?

1

2.65 λ

2

7.35 λ

3

5.47 λ

4

4.74 λ

Explanation:

Transition of electrons from the third orbit to the first orbit,
$\frac{1}{λ} = R [\frac{1}{1^{2}} - \frac{1}{3^{2}}]$
$\Rightarrow λ = \frac{9}{8 R}$
Transition of electrons from the fourth orbit to the second orbit,
$\frac{1}{λ^{'}} = R [\frac{1}{2^{2}} - \frac{1}{4^{2}}] \Rightarrow λ^{'} = \frac{16}{3 R}$
Now, $\frac{λ^{'}}{λ} = \frac{16}{3 R} \times \frac{8 R}{9} = \frac{128}{27}$
$\Rightarrow λ^{'} = \frac{128}{27} λ = 4.74 λ$