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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, $h = 4 \times 10^{- 15} e V - s$ )

1

974 n m

2

99.3 n m

3

941 n m

4

94.1 n m

Explanation:

Energy of emitted photon is given by,
$Δ E = h v = \frac{h c}{λ}$
$E_{1} - E_{2} = \frac{h c}{λ}$
As, $E = \frac{- 13.6 Z^{2}}{n^{2}} [Z = 1$ for hydrogen atom $]$
$\Rightarrow \frac{- 13.6 {(1)}^{2}}{{(4)}^{2}} - (\frac{13.6 {(1)}^{2}}{{(1)}^{2}})$
$= \frac{(4 \times 10^{- 15}) (3 \times 10^{8})}{λ}$
$\Rightarrow \frac{1}{λ} = 1.0625 \times 10^{7} m^{- 1}$
$λ = 0.941 \times 10^{- 7} m$
$\Rightarrow λ = 94.1 n m$ .
So , correct option is (4).

JEE - 2023

PHXII12:ATOMS

356565 Match the Column-I and Column-II
Column I
Column II
A
$n = 5 t o n = 2$
P
Lyman series
B
$n = 8 t o n = 4$
Q
Brackett series
C
$n = 3 t o n = 1$
R
Paschen
D
$n = 4 t o n = 3$
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

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 $λ$ is emitted. The wavelength of photon produced during its transistion from $\frac{4 E}{3}$ level to $E$ is:
supporting img

1

\frac{λ}{3}

2

\frac{3 λ}{4}

3

\frac{4 λ}{3}

4

3 λ

Explanation:

$2 E - E = \frac{h c}{λ} \Rightarrow E = \frac{h c}{λ}$
$\begin{aligned} \frac{4 E}{3} - 3 = \frac{h c}{λ^{'}} \Rightarrow \frac{E}{3} = \frac{h c}{λ^{'}} \\ ∴ \frac{λ^{'}}{λ} = 3 \Rightarrow λ^{'} = 3 λ . \end{aligned}$

PHXII12:ATOMS

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
$\frac{1}{λ} = R (\frac{1}{2^{2}} - \frac{1}{n^{2}})$ 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 $λ_{1} a n d λ_{2}$ are the wavelength of the Second member of the Lyman and Paschen series respectively, then $λ_{1} : λ_{2}$ is

1

1 : 3

2

1 : 30

3

7 : 50

4

18 : 225

Explanation:

For second line of Lyman series, $n_{1} = 1$ and $n_{2} = 3$
$∴ \frac{1}{λ_{1}} = R (\frac{1}{1^{2}} - \frac{1}{3^{2}}) = R (1 - \frac{1}{9}) = \frac{8 R}{9}$
For second line of paschen series, $n_{1} = 3$ and $n_{2} = 5$
$∴ \frac{1}{λ_{2}} = R (\frac{1}{3^{2}} - \frac{1}{5^{2}}) = R (\frac{1}{9} - \frac{1}{25}) = \frac{16 R}{225}$
$∴ \frac{λ_{1}}{λ_{2}} = \frac{16 R}{225} \times \frac{9}{8 R} = \frac{18}{225}$
$λ_{1} : λ_{2} = 18 : 225$

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, $h = 4 \times 10^{- 15} e V - s$ )

1

974 n m

2

99.3 n m

3

941 n m

4

94.1 n m

Explanation:

Energy of emitted photon is given by,
$Δ E = h v = \frac{h c}{λ}$
$E_{1} - E_{2} = \frac{h c}{λ}$
As, $E = \frac{- 13.6 Z^{2}}{n^{2}} [Z = 1$ for hydrogen atom $]$
$\Rightarrow \frac{- 13.6 {(1)}^{2}}{{(4)}^{2}} - (\frac{13.6 {(1)}^{2}}{{(1)}^{2}})$
$= \frac{(4 \times 10^{- 15}) (3 \times 10^{8})}{λ}$
$\Rightarrow \frac{1}{λ} = 1.0625 \times 10^{7} m^{- 1}$
$λ = 0.941 \times 10^{- 7} m$
$\Rightarrow λ = 94.1 n m$ .
So , correct option is (4).

JEE - 2023

PHXII12:ATOMS

356565 Match the Column-I and Column-II
Column I
Column II
A
$n = 5 t o n = 2$
P
Lyman series
B
$n = 8 t o n = 4$
Q
Brackett series
C
$n = 3 t o n = 1$
R
Paschen
D
$n = 4 t o n = 3$
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

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 $λ$ is emitted. The wavelength of photon produced during its transistion from $\frac{4 E}{3}$ level to $E$ is:
supporting img

1

\frac{λ}{3}

2

\frac{3 λ}{4}

3

\frac{4 λ}{3}

4

3 λ

Explanation:

$2 E - E = \frac{h c}{λ} \Rightarrow E = \frac{h c}{λ}$
$\begin{aligned} \frac{4 E}{3} - 3 = \frac{h c}{λ^{'}} \Rightarrow \frac{E}{3} = \frac{h c}{λ^{'}} \\ ∴ \frac{λ^{'}}{λ} = 3 \Rightarrow λ^{'} = 3 λ . \end{aligned}$

PHXII12:ATOMS

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
$\frac{1}{λ} = R (\frac{1}{2^{2}} - \frac{1}{n^{2}})$ 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 $λ_{1} a n d λ_{2}$ are the wavelength of the Second member of the Lyman and Paschen series respectively, then $λ_{1} : λ_{2}$ is

1

1 : 3

2

1 : 30

3

7 : 50

4

18 : 225

Explanation:

For second line of Lyman series, $n_{1} = 1$ and $n_{2} = 3$
$∴ \frac{1}{λ_{1}} = R (\frac{1}{1^{2}} - \frac{1}{3^{2}}) = R (1 - \frac{1}{9}) = \frac{8 R}{9}$
For second line of paschen series, $n_{1} = 3$ and $n_{2} = 5$
$∴ \frac{1}{λ_{2}} = R (\frac{1}{3^{2}} - \frac{1}{5^{2}}) = R (\frac{1}{9} - \frac{1}{25}) = \frac{16 R}{225}$
$∴ \frac{λ_{1}}{λ_{2}} = \frac{16 R}{225} \times \frac{9}{8 R} = \frac{18}{225}$
$λ_{1} : λ_{2} = 18 : 225$

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, $h = 4 \times 10^{- 15} e V - s$ )

1

974 n m

2

99.3 n m

3

941 n m

4

94.1 n m

Explanation:

Energy of emitted photon is given by,
$Δ E = h v = \frac{h c}{λ}$
$E_{1} - E_{2} = \frac{h c}{λ}$
As, $E = \frac{- 13.6 Z^{2}}{n^{2}} [Z = 1$ for hydrogen atom $]$
$\Rightarrow \frac{- 13.6 {(1)}^{2}}{{(4)}^{2}} - (\frac{13.6 {(1)}^{2}}{{(1)}^{2}})$
$= \frac{(4 \times 10^{- 15}) (3 \times 10^{8})}{λ}$
$\Rightarrow \frac{1}{λ} = 1.0625 \times 10^{7} m^{- 1}$
$λ = 0.941 \times 10^{- 7} m$
$\Rightarrow λ = 94.1 n m$ .
So , correct option is (4).

JEE - 2023

PHXII12:ATOMS

356565 Match the Column-I and Column-II
Column I
Column II
A
$n = 5 t o n = 2$
P
Lyman series
B
$n = 8 t o n = 4$
Q
Brackett series
C
$n = 3 t o n = 1$
R
Paschen
D
$n = 4 t o n = 3$
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

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 $λ$ is emitted. The wavelength of photon produced during its transistion from $\frac{4 E}{3}$ level to $E$ is:
supporting img

1

\frac{λ}{3}

2

\frac{3 λ}{4}

3

\frac{4 λ}{3}

4

3 λ

Explanation:

$2 E - E = \frac{h c}{λ} \Rightarrow E = \frac{h c}{λ}$
$\begin{aligned} \frac{4 E}{3} - 3 = \frac{h c}{λ^{'}} \Rightarrow \frac{E}{3} = \frac{h c}{λ^{'}} \\ ∴ \frac{λ^{'}}{λ} = 3 \Rightarrow λ^{'} = 3 λ . \end{aligned}$

PHXII12:ATOMS

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
$\frac{1}{λ} = R (\frac{1}{2^{2}} - \frac{1}{n^{2}})$ 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 $λ_{1} a n d λ_{2}$ are the wavelength of the Second member of the Lyman and Paschen series respectively, then $λ_{1} : λ_{2}$ is

1

1 : 3

2

1 : 30

3

7 : 50

4

18 : 225

Explanation:

For second line of Lyman series, $n_{1} = 1$ and $n_{2} = 3$
$∴ \frac{1}{λ_{1}} = R (\frac{1}{1^{2}} - \frac{1}{3^{2}}) = R (1 - \frac{1}{9}) = \frac{8 R}{9}$
For second line of paschen series, $n_{1} = 3$ and $n_{2} = 5$
$∴ \frac{1}{λ_{2}} = R (\frac{1}{3^{2}} - \frac{1}{5^{2}}) = R (\frac{1}{9} - \frac{1}{25}) = \frac{16 R}{225}$
$∴ \frac{λ_{1}}{λ_{2}} = \frac{16 R}{225} \times \frac{9}{8 R} = \frac{18}{225}$
$λ_{1} : λ_{2} = 18 : 225$

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, $h = 4 \times 10^{- 15} e V - s$ )

1

974 n m

2

99.3 n m

3

941 n m

4

94.1 n m

Explanation:

Energy of emitted photon is given by,
$Δ E = h v = \frac{h c}{λ}$
$E_{1} - E_{2} = \frac{h c}{λ}$
As, $E = \frac{- 13.6 Z^{2}}{n^{2}} [Z = 1$ for hydrogen atom $]$
$\Rightarrow \frac{- 13.6 {(1)}^{2}}{{(4)}^{2}} - (\frac{13.6 {(1)}^{2}}{{(1)}^{2}})$
$= \frac{(4 \times 10^{- 15}) (3 \times 10^{8})}{λ}$
$\Rightarrow \frac{1}{λ} = 1.0625 \times 10^{7} m^{- 1}$
$λ = 0.941 \times 10^{- 7} m$
$\Rightarrow λ = 94.1 n m$ .
So , correct option is (4).

JEE - 2023

PHXII12:ATOMS

356565 Match the Column-I and Column-II
Column I
Column II
A
$n = 5 t o n = 2$
P
Lyman series
B
$n = 8 t o n = 4$
Q
Brackett series
C
$n = 3 t o n = 1$
R
Paschen
D
$n = 4 t o n = 3$
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

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 $λ$ is emitted. The wavelength of photon produced during its transistion from $\frac{4 E}{3}$ level to $E$ is:
supporting img

1

\frac{λ}{3}

2

\frac{3 λ}{4}

3

\frac{4 λ}{3}

4

3 λ

Explanation:

$2 E - E = \frac{h c}{λ} \Rightarrow E = \frac{h c}{λ}$
$\begin{aligned} \frac{4 E}{3} - 3 = \frac{h c}{λ^{'}} \Rightarrow \frac{E}{3} = \frac{h c}{λ^{'}} \\ ∴ \frac{λ^{'}}{λ} = 3 \Rightarrow λ^{'} = 3 λ . \end{aligned}$

PHXII12:ATOMS

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
$\frac{1}{λ} = R (\frac{1}{2^{2}} - \frac{1}{n^{2}})$ 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 $λ_{1} a n d λ_{2}$ are the wavelength of the Second member of the Lyman and Paschen series respectively, then $λ_{1} : λ_{2}$ is

1

1 : 3

2

1 : 30

3

7 : 50

4

18 : 225

Explanation:

For second line of Lyman series, $n_{1} = 1$ and $n_{2} = 3$
$∴ \frac{1}{λ_{1}} = R (\frac{1}{1^{2}} - \frac{1}{3^{2}}) = R (1 - \frac{1}{9}) = \frac{8 R}{9}$
For second line of paschen series, $n_{1} = 3$ and $n_{2} = 5$
$∴ \frac{1}{λ_{2}} = R (\frac{1}{3^{2}} - \frac{1}{5^{2}}) = R (\frac{1}{9} - \frac{1}{25}) = \frac{16 R}{225}$
$∴ \frac{λ_{1}}{λ_{2}} = \frac{16 R}{225} \times \frac{9}{8 R} = \frac{18}{225}$
$λ_{1} : λ_{2} = 18 : 225$

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, $h = 4 \times 10^{- 15} e V - s$ )

1

974 n m

2

99.3 n m

3

941 n m

4

94.1 n m

Explanation:

Energy of emitted photon is given by,
$Δ E = h v = \frac{h c}{λ}$
$E_{1} - E_{2} = \frac{h c}{λ}$
As, $E = \frac{- 13.6 Z^{2}}{n^{2}} [Z = 1$ for hydrogen atom $]$
$\Rightarrow \frac{- 13.6 {(1)}^{2}}{{(4)}^{2}} - (\frac{13.6 {(1)}^{2}}{{(1)}^{2}})$
$= \frac{(4 \times 10^{- 15}) (3 \times 10^{8})}{λ}$
$\Rightarrow \frac{1}{λ} = 1.0625 \times 10^{7} m^{- 1}$
$λ = 0.941 \times 10^{- 7} m$
$\Rightarrow λ = 94.1 n m$ .
So , correct option is (4).

JEE - 2023

PHXII12:ATOMS

356565 Match the Column-I and Column-II
Column I
Column II
A
$n = 5 t o n = 2$
P
Lyman series
B
$n = 8 t o n = 4$
Q
Brackett series
C
$n = 3 t o n = 1$
R
Paschen
D
$n = 4 t o n = 3$
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

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 $λ$ is emitted. The wavelength of photon produced during its transistion from $\frac{4 E}{3}$ level to $E$ is:
supporting img

1

\frac{λ}{3}

2

\frac{3 λ}{4}

3

\frac{4 λ}{3}

4

3 λ

Explanation:

$2 E - E = \frac{h c}{λ} \Rightarrow E = \frac{h c}{λ}$
$\begin{aligned} \frac{4 E}{3} - 3 = \frac{h c}{λ^{'}} \Rightarrow \frac{E}{3} = \frac{h c}{λ^{'}} \\ ∴ \frac{λ^{'}}{λ} = 3 \Rightarrow λ^{'} = 3 λ . \end{aligned}$

PHXII12:ATOMS

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
$\frac{1}{λ} = R (\frac{1}{2^{2}} - \frac{1}{n^{2}})$ 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 $λ_{1} a n d λ_{2}$ are the wavelength of the Second member of the Lyman and Paschen series respectively, then $λ_{1} : λ_{2}$ is

1

1 : 3

2

1 : 30

3

7 : 50

4

18 : 225

Explanation:

For second line of Lyman series, $n_{1} = 1$ and $n_{2} = 3$
$∴ \frac{1}{λ_{1}} = R (\frac{1}{1^{2}} - \frac{1}{3^{2}}) = R (1 - \frac{1}{9}) = \frac{8 R}{9}$
For second line of paschen series, $n_{1} = 3$ and $n_{2} = 5$
$∴ \frac{1}{λ_{2}} = R (\frac{1}{3^{2}} - \frac{1}{5^{2}}) = R (\frac{1}{9} - \frac{1}{25}) = \frac{16 R}{225}$
$∴ \frac{λ_{1}}{λ_{2}} = \frac{16 R}{225} \times \frac{9}{8 R} = \frac{18}{225}$
$λ_{1} : λ_{2} = 18 : 225$

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