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PHXII12:ATOMS

356620 Which of the following spectral series of hydrogen atom is lying in visible range of electromagnetic wave?

1 Lyman series

2 Paschen series

3 Balmer series

4 Pfund series

Explanation:

The Balmer series lies in the visible region. The Lyman series is in the ultraviolet region, and Paschen and Pfund series are in the infrared region.
$v = \frac{h}{m λ} = \frac{24}{25} \frac{R h}{m}$

KCET - 2014

PHXII12:ATOMS

356621 An electron jumps from $4^{t h}$ orbit to $2^{n d}$ orbit of the hydrogen atom. Given, Rydberg's constant $R = 10^{5} c m^{- 1}$ , the frequency in hertz of emitted radiation will be

1

\frac{3}{16} \times 10^{5}

2

\frac{3}{16} \times 10^{15}

3

\frac{9}{16} \times 10^{15}

4

\frac{3}{4} \times 10^{15}

Explanation:

We know that, frequency,
$v = \frac{c}{λ} = c \cdot R (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$
$= 3 \times 10^{8} \times 10^{7} (\frac{1}{2^{2}} - \frac{1}{4^{2}})$
$= \frac{9}{16} \times 10^{15} H z$

PHXII12:ATOMS

356622 The figure shows the energy level of certain atom. When the electron deexcites from $3 E$ to $E$ , an electromagnetic wave of wavelength $λ$ is emitted. What is the wavelength of the electromagnetic wave emitted when the electron deexcites from $5 E / 3$ to $E$ ?
supporting img

1

3 λ

2

2 λ

3

5 λ

4

3 λ / 5

Explanation:

From figure,
$\frac{h c}{λ} = 3 E - E = 2 E (1)$
$\frac{h c}{λ^{1}} = \frac{5 E}{3} - E = \frac{2}{3} E (2)$
Divide (1) by (2), we get
$\frac{λ^{'}}{λ} = 3$ or $λ^{'} = 3 λ$
supporting img

KCET - 2013

PHXII12:ATOMS

356623 Electron in hydrogen atom first jumps from third excited state to second excited state and then from second excited to the first excited state. The ratio of the wavelengths $λ_{1} : λ_{2}$ emitted in the two cases is

1

20 / 7

2

7 / 5

3

27 / 20

4

27 / 5

Explanation:

We know that, $\frac{1}{λ} = R [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$
So, for $λ_{1}$
$\Rightarrow \frac{1}{λ_{1}} = R [\frac{1}{{(3)}^{2}} - \frac{1}{{(4)}^{2}}]$
$\frac{1}{λ_{1}} = R [\frac{7}{144}]$ (1)
Similarly, for $λ_{2}$
$\Rightarrow \frac{1}{λ_{2}} = R [\frac{1}{{(2)}^{2}} - \frac{1}{{(3)}^{2}}]$
$\frac{1}{λ_{2}} = R [\frac{5}{36}]$ (2)
Hence, from Eqs. (1) and (2), we get
$\frac{λ_{1}}{λ_{2}} = \frac{20}{7}$

PHXII12:ATOMS

356620 Which of the following spectral series of hydrogen atom is lying in visible range of electromagnetic wave?

1 Lyman series

2 Paschen series

3 Balmer series

4 Pfund series

Explanation:

The Balmer series lies in the visible region. The Lyman series is in the ultraviolet region, and Paschen and Pfund series are in the infrared region.
$v = \frac{h}{m λ} = \frac{24}{25} \frac{R h}{m}$

KCET - 2014

PHXII12:ATOMS

356621 An electron jumps from $4^{t h}$ orbit to $2^{n d}$ orbit of the hydrogen atom. Given, Rydberg's constant $R = 10^{5} c m^{- 1}$ , the frequency in hertz of emitted radiation will be

1

\frac{3}{16} \times 10^{5}

2

\frac{3}{16} \times 10^{15}

3

\frac{9}{16} \times 10^{15}

4

\frac{3}{4} \times 10^{15}

Explanation:

We know that, frequency,
$v = \frac{c}{λ} = c \cdot R (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$
$= 3 \times 10^{8} \times 10^{7} (\frac{1}{2^{2}} - \frac{1}{4^{2}})$
$= \frac{9}{16} \times 10^{15} H z$

PHXII12:ATOMS

356622 The figure shows the energy level of certain atom. When the electron deexcites from $3 E$ to $E$ , an electromagnetic wave of wavelength $λ$ is emitted. What is the wavelength of the electromagnetic wave emitted when the electron deexcites from $5 E / 3$ to $E$ ?
supporting img

1

3 λ

2

2 λ

3

5 λ

4

3 λ / 5

Explanation:

From figure,
$\frac{h c}{λ} = 3 E - E = 2 E (1)$
$\frac{h c}{λ^{1}} = \frac{5 E}{3} - E = \frac{2}{3} E (2)$
Divide (1) by (2), we get
$\frac{λ^{'}}{λ} = 3$ or $λ^{'} = 3 λ$
supporting img

KCET - 2013

PHXII12:ATOMS

356623 Electron in hydrogen atom first jumps from third excited state to second excited state and then from second excited to the first excited state. The ratio of the wavelengths $λ_{1} : λ_{2}$ emitted in the two cases is

1

20 / 7

2

7 / 5

3

27 / 20

4

27 / 5

Explanation:

We know that, $\frac{1}{λ} = R [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$
So, for $λ_{1}$
$\Rightarrow \frac{1}{λ_{1}} = R [\frac{1}{{(3)}^{2}} - \frac{1}{{(4)}^{2}}]$
$\frac{1}{λ_{1}} = R [\frac{7}{144}]$ (1)
Similarly, for $λ_{2}$
$\Rightarrow \frac{1}{λ_{2}} = R [\frac{1}{{(2)}^{2}} - \frac{1}{{(3)}^{2}}]$
$\frac{1}{λ_{2}} = R [\frac{5}{36}]$ (2)
Hence, from Eqs. (1) and (2), we get
$\frac{λ_{1}}{λ_{2}} = \frac{20}{7}$

PHXII12:ATOMS

356620 Which of the following spectral series of hydrogen atom is lying in visible range of electromagnetic wave?

1 Lyman series

2 Paschen series

3 Balmer series

4 Pfund series

Explanation:

The Balmer series lies in the visible region. The Lyman series is in the ultraviolet region, and Paschen and Pfund series are in the infrared region.
$v = \frac{h}{m λ} = \frac{24}{25} \frac{R h}{m}$

KCET - 2014

PHXII12:ATOMS

356621 An electron jumps from $4^{t h}$ orbit to $2^{n d}$ orbit of the hydrogen atom. Given, Rydberg's constant $R = 10^{5} c m^{- 1}$ , the frequency in hertz of emitted radiation will be

1

\frac{3}{16} \times 10^{5}

2

\frac{3}{16} \times 10^{15}

3

\frac{9}{16} \times 10^{15}

4

\frac{3}{4} \times 10^{15}

Explanation:

We know that, frequency,
$v = \frac{c}{λ} = c \cdot R (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$
$= 3 \times 10^{8} \times 10^{7} (\frac{1}{2^{2}} - \frac{1}{4^{2}})$
$= \frac{9}{16} \times 10^{15} H z$

PHXII12:ATOMS

356622 The figure shows the energy level of certain atom. When the electron deexcites from $3 E$ to $E$ , an electromagnetic wave of wavelength $λ$ is emitted. What is the wavelength of the electromagnetic wave emitted when the electron deexcites from $5 E / 3$ to $E$ ?
supporting img

1

3 λ

2

2 λ

3

5 λ

4

3 λ / 5

Explanation:

From figure,
$\frac{h c}{λ} = 3 E - E = 2 E (1)$
$\frac{h c}{λ^{1}} = \frac{5 E}{3} - E = \frac{2}{3} E (2)$
Divide (1) by (2), we get
$\frac{λ^{'}}{λ} = 3$ or $λ^{'} = 3 λ$
supporting img

KCET - 2013

PHXII12:ATOMS

356623 Electron in hydrogen atom first jumps from third excited state to second excited state and then from second excited to the first excited state. The ratio of the wavelengths $λ_{1} : λ_{2}$ emitted in the two cases is

1

20 / 7

2

7 / 5

3

27 / 20

4

27 / 5

Explanation:

We know that, $\frac{1}{λ} = R [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$
So, for $λ_{1}$
$\Rightarrow \frac{1}{λ_{1}} = R [\frac{1}{{(3)}^{2}} - \frac{1}{{(4)}^{2}}]$
$\frac{1}{λ_{1}} = R [\frac{7}{144}]$ (1)
Similarly, for $λ_{2}$
$\Rightarrow \frac{1}{λ_{2}} = R [\frac{1}{{(2)}^{2}} - \frac{1}{{(3)}^{2}}]$
$\frac{1}{λ_{2}} = R [\frac{5}{36}]$ (2)
Hence, from Eqs. (1) and (2), we get
$\frac{λ_{1}}{λ_{2}} = \frac{20}{7}$

PHXII12:ATOMS

356620 Which of the following spectral series of hydrogen atom is lying in visible range of electromagnetic wave?

1 Lyman series

2 Paschen series

3 Balmer series

4 Pfund series

Explanation:

The Balmer series lies in the visible region. The Lyman series is in the ultraviolet region, and Paschen and Pfund series are in the infrared region.
$v = \frac{h}{m λ} = \frac{24}{25} \frac{R h}{m}$

KCET - 2014

PHXII12:ATOMS

356621 An electron jumps from $4^{t h}$ orbit to $2^{n d}$ orbit of the hydrogen atom. Given, Rydberg's constant $R = 10^{5} c m^{- 1}$ , the frequency in hertz of emitted radiation will be

1

\frac{3}{16} \times 10^{5}

2

\frac{3}{16} \times 10^{15}

3

\frac{9}{16} \times 10^{15}

4

\frac{3}{4} \times 10^{15}

Explanation:

We know that, frequency,
$v = \frac{c}{λ} = c \cdot R (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$
$= 3 \times 10^{8} \times 10^{7} (\frac{1}{2^{2}} - \frac{1}{4^{2}})$
$= \frac{9}{16} \times 10^{15} H z$

PHXII12:ATOMS

356622 The figure shows the energy level of certain atom. When the electron deexcites from $3 E$ to $E$ , an electromagnetic wave of wavelength $λ$ is emitted. What is the wavelength of the electromagnetic wave emitted when the electron deexcites from $5 E / 3$ to $E$ ?
supporting img

1

3 λ

2

2 λ

3

5 λ

4

3 λ / 5

Explanation:

From figure,
$\frac{h c}{λ} = 3 E - E = 2 E (1)$
$\frac{h c}{λ^{1}} = \frac{5 E}{3} - E = \frac{2}{3} E (2)$
Divide (1) by (2), we get
$\frac{λ^{'}}{λ} = 3$ or $λ^{'} = 3 λ$
supporting img

KCET - 2013

PHXII12:ATOMS

356623 Electron in hydrogen atom first jumps from third excited state to second excited state and then from second excited to the first excited state. The ratio of the wavelengths $λ_{1} : λ_{2}$ emitted in the two cases is

1

20 / 7

2

7 / 5

3

27 / 20

4

27 / 5

Explanation:

We know that, $\frac{1}{λ} = R [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$
So, for $λ_{1}$
$\Rightarrow \frac{1}{λ_{1}} = R [\frac{1}{{(3)}^{2}} - \frac{1}{{(4)}^{2}}]$
$\frac{1}{λ_{1}} = R [\frac{7}{144}]$ (1)
Similarly, for $λ_{2}$
$\Rightarrow \frac{1}{λ_{2}} = R [\frac{1}{{(2)}^{2}} - \frac{1}{{(3)}^{2}}]$
$\frac{1}{λ_{2}} = R [\frac{5}{36}]$ (2)
Hence, from Eqs. (1) and (2), we get
$\frac{λ_{1}}{λ_{2}} = \frac{20}{7}$

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