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CHXI02:STRUCTURE OF ATOM

307574 Electronic transition in $H e^{+}$ ion takes from $n_{2}$ to $n_{1}$ shell when electron transition to $1^{s t}$ shell from 5th shell, the number of spectral line formed are

1

21

2

15

3

20

4

10

Explanation:

Number of lines $= \frac{n (n + 1)}{2}$
where, $n = n_{2} - n_{1} = 5 - 1 = 4 .$
No. of lines $\frac{4 \times 5}{2} = 10 .$

CHXI02:STRUCTURE OF ATOM

307575 Which transition in the hydrogen atomic spectrum will have the same wavelength as the Balmer transition (i.e. $n = 4$ to $n = 2$ ) of ${He}^{+}$ spectrum?

1

n = 4

to

n = 3

2

n = 3

to

n = 2

3

n = 4

to

n = 2

4

n = 2

to

n = 1

Explanation:

$∵ \frac{1}{λ} = Z^{2} \cdot R [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$
$∴$ For ${He}^{+}$ ion, $Z = 2, n_{1} = 2, n_{2} = 4$
$\frac{1}{λ} = 2^{2} .R [\frac{1}{2^{2}} - \frac{1}{4^{2}}]$
$\frac{1}{λ} = 4 \times R \times \frac{3}{16} = \frac{3}{4} R$
The same value of for $H$ -atom is possible when electron jumps from $n = 2$ to $n = 1$ ,
i.e. $\frac{1}{λ} = 1^{2} \times R [\frac{1}{1^{2}} - \frac{1}{2^{2}}]$
$\Rightarrow= 1 R [\frac{1}{1} - \frac{1}{4}] = \frac{3}{4} R$

AIIMS - 2017

CHXI02:STRUCTURE OF ATOM

307576 The ionization enthalpy of hydrogen atom is $1 .312 \times 1 0^{6} J / m o l$ . The energy required to excite the elctron in the atom from to $n = 1 t o n = 2$ is

1

8 .51 \times 1 0^{5} J / m o l

2

6 .56 \times 1 0^{5} J / m o l

3

7 .56 \times 1 0^{5} J / m o l

4

9 .84 \times 1 0^{5} J / m o l

Explanation:

$Δ E = E_{2} - E_{1} = - \frac{E_{1}}{2^{2}} + \frac{E_{1}}{1^{2}}$
$= \frac{1 .312 \times 1 0^{6}}{2^{2}} + 1 .312 \times 1 0^{6} = 9 .84 \times 1 0^{5} J / m o l$

CHXI02:STRUCTURE OF ATOM

307577 Monochromatic radiation of specific wavelength is incident on H-atoms in ground state. H-atoms absorb energy and emit subsequently radiations of six different wavelength. Find wavelength of incident radiations:

1

9 .75 n m

2

50 n m

3

85 .8 n m

4

97 .25 n m

Explanation:

$\frac{n (n - 1)}{2} = 6; n = 4$
$n = 4, E_{4} = - 0 . 85 e V$
$n = 1, E_{1} = - 13 . 6 e v$
$∴ Δ E = 12 . 75 e V$
$12 .75 e V = \frac{1240 e V - n m}{λ}$
$λ = 97 .25 n m$

CHXI02:STRUCTURE OF ATOM

307574 Electronic transition in $H e^{+}$ ion takes from $n_{2}$ to $n_{1}$ shell when electron transition to $1^{s t}$ shell from 5th shell, the number of spectral line formed are

1

21

2

15

3

20

4

10

Explanation:

Number of lines $= \frac{n (n + 1)}{2}$
where, $n = n_{2} - n_{1} = 5 - 1 = 4 .$
No. of lines $\frac{4 \times 5}{2} = 10 .$

CHXI02:STRUCTURE OF ATOM

307575 Which transition in the hydrogen atomic spectrum will have the same wavelength as the Balmer transition (i.e. $n = 4$ to $n = 2$ ) of ${He}^{+}$ spectrum?

1

n = 4

to

n = 3

2

n = 3

to

n = 2

3

n = 4

to

n = 2

4

n = 2

to

n = 1

Explanation:

$∵ \frac{1}{λ} = Z^{2} \cdot R [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$
$∴$ For ${He}^{+}$ ion, $Z = 2, n_{1} = 2, n_{2} = 4$
$\frac{1}{λ} = 2^{2} .R [\frac{1}{2^{2}} - \frac{1}{4^{2}}]$
$\frac{1}{λ} = 4 \times R \times \frac{3}{16} = \frac{3}{4} R$
The same value of for $H$ -atom is possible when electron jumps from $n = 2$ to $n = 1$ ,
i.e. $\frac{1}{λ} = 1^{2} \times R [\frac{1}{1^{2}} - \frac{1}{2^{2}}]$
$\Rightarrow= 1 R [\frac{1}{1} - \frac{1}{4}] = \frac{3}{4} R$

AIIMS - 2017

CHXI02:STRUCTURE OF ATOM

307576 The ionization enthalpy of hydrogen atom is $1 .312 \times 1 0^{6} J / m o l$ . The energy required to excite the elctron in the atom from to $n = 1 t o n = 2$ is

1

8 .51 \times 1 0^{5} J / m o l

2

6 .56 \times 1 0^{5} J / m o l

3

7 .56 \times 1 0^{5} J / m o l

4

9 .84 \times 1 0^{5} J / m o l

Explanation:

$Δ E = E_{2} - E_{1} = - \frac{E_{1}}{2^{2}} + \frac{E_{1}}{1^{2}}$
$= \frac{1 .312 \times 1 0^{6}}{2^{2}} + 1 .312 \times 1 0^{6} = 9 .84 \times 1 0^{5} J / m o l$

CHXI02:STRUCTURE OF ATOM

307577 Monochromatic radiation of specific wavelength is incident on H-atoms in ground state. H-atoms absorb energy and emit subsequently radiations of six different wavelength. Find wavelength of incident radiations:

1

9 .75 n m

2

50 n m

3

85 .8 n m

4

97 .25 n m

Explanation:

$\frac{n (n - 1)}{2} = 6; n = 4$
$n = 4, E_{4} = - 0 . 85 e V$
$n = 1, E_{1} = - 13 . 6 e v$
$∴ Δ E = 12 . 75 e V$
$12 .75 e V = \frac{1240 e V - n m}{λ}$
$λ = 97 .25 n m$

CHXI02:STRUCTURE OF ATOM

307574 Electronic transition in $H e^{+}$ ion takes from $n_{2}$ to $n_{1}$ shell when electron transition to $1^{s t}$ shell from 5th shell, the number of spectral line formed are

1

21

2

15

3

20

4

10

Explanation:

Number of lines $= \frac{n (n + 1)}{2}$
where, $n = n_{2} - n_{1} = 5 - 1 = 4 .$
No. of lines $\frac{4 \times 5}{2} = 10 .$

CHXI02:STRUCTURE OF ATOM

307575 Which transition in the hydrogen atomic spectrum will have the same wavelength as the Balmer transition (i.e. $n = 4$ to $n = 2$ ) of ${He}^{+}$ spectrum?

1

n = 4

to

n = 3

2

n = 3

to

n = 2

3

n = 4

to

n = 2

4

n = 2

to

n = 1

Explanation:

$∵ \frac{1}{λ} = Z^{2} \cdot R [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$
$∴$ For ${He}^{+}$ ion, $Z = 2, n_{1} = 2, n_{2} = 4$
$\frac{1}{λ} = 2^{2} .R [\frac{1}{2^{2}} - \frac{1}{4^{2}}]$
$\frac{1}{λ} = 4 \times R \times \frac{3}{16} = \frac{3}{4} R$
The same value of for $H$ -atom is possible when electron jumps from $n = 2$ to $n = 1$ ,
i.e. $\frac{1}{λ} = 1^{2} \times R [\frac{1}{1^{2}} - \frac{1}{2^{2}}]$
$\Rightarrow= 1 R [\frac{1}{1} - \frac{1}{4}] = \frac{3}{4} R$

AIIMS - 2017

CHXI02:STRUCTURE OF ATOM

307576 The ionization enthalpy of hydrogen atom is $1 .312 \times 1 0^{6} J / m o l$ . The energy required to excite the elctron in the atom from to $n = 1 t o n = 2$ is

1

8 .51 \times 1 0^{5} J / m o l

2

6 .56 \times 1 0^{5} J / m o l

3

7 .56 \times 1 0^{5} J / m o l

4

9 .84 \times 1 0^{5} J / m o l

Explanation:

$Δ E = E_{2} - E_{1} = - \frac{E_{1}}{2^{2}} + \frac{E_{1}}{1^{2}}$
$= \frac{1 .312 \times 1 0^{6}}{2^{2}} + 1 .312 \times 1 0^{6} = 9 .84 \times 1 0^{5} J / m o l$

CHXI02:STRUCTURE OF ATOM

307577 Monochromatic radiation of specific wavelength is incident on H-atoms in ground state. H-atoms absorb energy and emit subsequently radiations of six different wavelength. Find wavelength of incident radiations:

1

9 .75 n m

2

50 n m

3

85 .8 n m

4

97 .25 n m

Explanation:

$\frac{n (n - 1)}{2} = 6; n = 4$
$n = 4, E_{4} = - 0 . 85 e V$
$n = 1, E_{1} = - 13 . 6 e v$
$∴ Δ E = 12 . 75 e V$
$12 .75 e V = \frac{1240 e V - n m}{λ}$
$λ = 97 .25 n m$

CHXI02:STRUCTURE OF ATOM

307574 Electronic transition in $H e^{+}$ ion takes from $n_{2}$ to $n_{1}$ shell when electron transition to $1^{s t}$ shell from 5th shell, the number of spectral line formed are

1

21

2

15

3

20

4

10

Explanation:

Number of lines $= \frac{n (n + 1)}{2}$
where, $n = n_{2} - n_{1} = 5 - 1 = 4 .$
No. of lines $\frac{4 \times 5}{2} = 10 .$

CHXI02:STRUCTURE OF ATOM

307575 Which transition in the hydrogen atomic spectrum will have the same wavelength as the Balmer transition (i.e. $n = 4$ to $n = 2$ ) of ${He}^{+}$ spectrum?

1

n = 4

to

n = 3

2

n = 3

to

n = 2

3

n = 4

to

n = 2

4

n = 2

to

n = 1

Explanation:

$∵ \frac{1}{λ} = Z^{2} \cdot R [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$
$∴$ For ${He}^{+}$ ion, $Z = 2, n_{1} = 2, n_{2} = 4$
$\frac{1}{λ} = 2^{2} .R [\frac{1}{2^{2}} - \frac{1}{4^{2}}]$
$\frac{1}{λ} = 4 \times R \times \frac{3}{16} = \frac{3}{4} R$
The same value of for $H$ -atom is possible when electron jumps from $n = 2$ to $n = 1$ ,
i.e. $\frac{1}{λ} = 1^{2} \times R [\frac{1}{1^{2}} - \frac{1}{2^{2}}]$
$\Rightarrow= 1 R [\frac{1}{1} - \frac{1}{4}] = \frac{3}{4} R$

AIIMS - 2017

CHXI02:STRUCTURE OF ATOM

307576 The ionization enthalpy of hydrogen atom is $1 .312 \times 1 0^{6} J / m o l$ . The energy required to excite the elctron in the atom from to $n = 1 t o n = 2$ is

1

8 .51 \times 1 0^{5} J / m o l

2

6 .56 \times 1 0^{5} J / m o l

3

7 .56 \times 1 0^{5} J / m o l

4

9 .84 \times 1 0^{5} J / m o l

Explanation:

$Δ E = E_{2} - E_{1} = - \frac{E_{1}}{2^{2}} + \frac{E_{1}}{1^{2}}$
$= \frac{1 .312 \times 1 0^{6}}{2^{2}} + 1 .312 \times 1 0^{6} = 9 .84 \times 1 0^{5} J / m o l$

CHXI02:STRUCTURE OF ATOM

307577 Monochromatic radiation of specific wavelength is incident on H-atoms in ground state. H-atoms absorb energy and emit subsequently radiations of six different wavelength. Find wavelength of incident radiations:

1

9 .75 n m

2

50 n m

3

85 .8 n m

4

97 .25 n m

Explanation:

$\frac{n (n - 1)}{2} = 6; n = 4$
$n = 4, E_{4} = - 0 . 85 e V$
$n = 1, E_{1} = - 13 . 6 e v$
$∴ Δ E = 12 . 75 e V$
$12 .75 e V = \frac{1240 e V - n m}{λ}$
$λ = 97 .25 n m$