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ATOMS

145559 The second line of Balmer series has wavelength $4861 \AA$ . The wavelength of the first line of Balmer series is

1

1216 \AA

2

6563 \AA

3

4340 \AA

4

4101 \AA

Explanation:

B For the first line Balmer series
$\frac{1}{λ} = R (\frac{1}{2^{2}} - \frac{1}{3^{2}}) = \frac{5 R}{36}$
For second line Balmer series.
$\frac{1}{4861} = R (\frac{1}{2^{2}} - \frac{1}{4^{2}}) = \frac{3 R}{16}$
divide equation (ii) by (i) then
$\frac{λ}{4861} = \frac{3 R}{16} \times \frac{36}{5 R}$
$λ = 4861 \times \frac{27}{20} = 6563 \AA$

AP EAMCET (18.09.2020) Shift-II

ATOMS

145561 If $λ_{1}$ and $λ_{2}$ are the wavelengths of the first members of the Lyman and Paschen series respectively, then $λ_{1} : λ_{2}$ is

1

1 : 3

2

1 : 30

3

7 : 50

4

7 : 108

Explanation:

D In the Lyman series, the first line is
$n_{1} = 1$ and $n_{2} = 2$
$\frac{1}{λ_{1}} = R [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$
$\frac{1}{λ_{1}} = R [1 - \frac{1}{2^{2}}]$
$\frac{1}{λ_{1}} = R [1 - \frac{1}{4}]$
$\frac{1}{λ_{1}} = \frac{3 R}{4}$
$λ_{1} = \frac{4}{3 R}$
Similarly in Paschen series the first line is
$n_{1} = 3 and n_{2} = 4$
$\frac{1}{λ_{2}} = R [\frac{1}{3^{2}} - \frac{1}{4^{2}}]$
$\frac{1}{λ_{2}} = R [\frac{1}{9} - \frac{1}{16}]$
$\frac{1}{λ_{2}} = \frac{7}{144} R$
$λ_{2} = \frac{144}{7 R}$
From (i) and (ii)
$\frac{λ_{1}}{λ_{2}} = \frac{4}{3 R} \times \frac{7 R}{144}$
$\frac{λ_{1}}{λ_{2}} = \frac{7}{108}$
$∴ λ_{1} : λ_{2} = 7 : 108$

MHT CET-2020

ATOMS

145562 A hydrogen like atom has one electron revolving round a stationary nucleus. If the energy required to excite the electron from the $2^{nd}$ orbital to $3^{rd}$ orbit is $47.2 eV$ , find the atomic number of the given atom.

1 3

2 4

3 5

4 6

Explanation:

C Given that, $Δ E = 47.2 eV, n_{1} = 2, n_{2} = 3$
We know that,
$Δ E = 13.6 Z^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$
$Δ E = 13.6 \times Z^{2} \times (\frac{1}{2^{2}} - \frac{1}{3^{2}})$
$47.2 = 13.6 Z^{2} \times \frac{5}{36}$
$Z^{2} = \frac{47.2}{13.6} \times \frac{36}{5} = 25$
$Z = 5$

AP EAMCET (18.09.2020) Shift-I

ATOMS

145563 In hydrogen spectrum, if the shortest wavelength in Balmer series is $λ$ , the shortest wavelength in Brackett series is

1

λ

2

λ / 2

3

4 λ

4

9 λ

Explanation:

C Shortest wavelength for electron transit from $\infty$ to $n$
$\frac{1}{λ} = R (\frac{1}{n^{2}})$
For Balmer series, $n = 2$
$\frac{1}{λ} = R (\frac{1}{2^{2}})$
$\frac{1}{λ} = \frac{R}{4} \Rightarrow R = \frac{4}{λ}$
For Brackett series, $n = 4$
$\frac{1}{λ (b)} = R (\frac{1}{n^{2}}) = \frac{R}{16}$
$\frac{1}{λ (b)} = \frac{R}{16}$
$\frac{16}{λ (b)} = R$
$\frac{16}{λ (b)} = \frac{4}{λ}$
$4 λ (b) = 16 λ$
$λ (b) = 4 λ$

TS- EAMCET-09.09.2020

ATOMS

145564 If the first line in the Lyman series has wavelength $λ$ , then the first line in Balmer series has the wavelength

1

\frac{27}{5} λ

2

\frac{32}{27} λ

3

\frac{28}{21} λ

4

\frac{15}{4} λ

Explanation:

A For hydrogen atom,
For first line in Lyman series
$n_{1} = 1, and n_{2} = 2$
$\frac{1}{λ_{L}} = R (\frac{1}{1^{2}} - \frac{1}{2^{2}})$
$\frac{1}{λ_{L}} = R (\frac{1}{1} - \frac{1}{4})$
$\frac{1}{λ_{L}} = \frac{3 R}{4}$
For first line in Balmer series $n_{1} = 2, and n_{2} = 3$
$\frac{1}{λ_{B}} = R (\frac{1}{2^{2}} - \frac{1}{3^{2}})$
$\frac{1}{λ_{B}} = R (\frac{1}{4} - \frac{1}{9})$
$\frac{1}{λ_{B}} = R (\frac{9 - 4}{36})$
$\frac{1}{λ_{B}} = \frac{5 R}{36}$
From equation (i) and equation (ii), we get-
$\frac{1}{λ_{B}} = \frac{5 R}{\frac{1}{36}}$
$\frac{λ_{L}}{4 R}$
$\frac{1 / λ_{B}}{1 / λ_{L}} = \frac{5 R}{36} \times \frac{4}{3 R}$
$\frac{λ_{L}}{λ_{B}} = \frac{5 R}{27 R}$
$λ_{B} = \frac{27}{5} λ_{l}$
$(λ_{L} = λ)$
$λ_{B} = \frac{27}{5} λ$

TS- EAMCET.14.09.2020

ATOMS

145559 The second line of Balmer series has wavelength $4861 \AA$ . The wavelength of the first line of Balmer series is

1

1216 \AA

2

6563 \AA

3

4340 \AA

4

4101 \AA

Explanation:

B For the first line Balmer series
$\frac{1}{λ} = R (\frac{1}{2^{2}} - \frac{1}{3^{2}}) = \frac{5 R}{36}$
For second line Balmer series.
$\frac{1}{4861} = R (\frac{1}{2^{2}} - \frac{1}{4^{2}}) = \frac{3 R}{16}$
divide equation (ii) by (i) then
$\frac{λ}{4861} = \frac{3 R}{16} \times \frac{36}{5 R}$
$λ = 4861 \times \frac{27}{20} = 6563 \AA$

AP EAMCET (18.09.2020) Shift-II

ATOMS

145561 If $λ_{1}$ and $λ_{2}$ are the wavelengths of the first members of the Lyman and Paschen series respectively, then $λ_{1} : λ_{2}$ is

1

1 : 3

2

1 : 30

3

7 : 50

4

7 : 108

Explanation:

D In the Lyman series, the first line is
$n_{1} = 1$ and $n_{2} = 2$
$\frac{1}{λ_{1}} = R [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$
$\frac{1}{λ_{1}} = R [1 - \frac{1}{2^{2}}]$
$\frac{1}{λ_{1}} = R [1 - \frac{1}{4}]$
$\frac{1}{λ_{1}} = \frac{3 R}{4}$
$λ_{1} = \frac{4}{3 R}$
Similarly in Paschen series the first line is
$n_{1} = 3 and n_{2} = 4$
$\frac{1}{λ_{2}} = R [\frac{1}{3^{2}} - \frac{1}{4^{2}}]$
$\frac{1}{λ_{2}} = R [\frac{1}{9} - \frac{1}{16}]$
$\frac{1}{λ_{2}} = \frac{7}{144} R$
$λ_{2} = \frac{144}{7 R}$
From (i) and (ii)
$\frac{λ_{1}}{λ_{2}} = \frac{4}{3 R} \times \frac{7 R}{144}$
$\frac{λ_{1}}{λ_{2}} = \frac{7}{108}$
$∴ λ_{1} : λ_{2} = 7 : 108$

MHT CET-2020

ATOMS

145562 A hydrogen like atom has one electron revolving round a stationary nucleus. If the energy required to excite the electron from the $2^{nd}$ orbital to $3^{rd}$ orbit is $47.2 eV$ , find the atomic number of the given atom.

1 3

2 4

3 5

4 6

Explanation:

C Given that, $Δ E = 47.2 eV, n_{1} = 2, n_{2} = 3$
We know that,
$Δ E = 13.6 Z^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$
$Δ E = 13.6 \times Z^{2} \times (\frac{1}{2^{2}} - \frac{1}{3^{2}})$
$47.2 = 13.6 Z^{2} \times \frac{5}{36}$
$Z^{2} = \frac{47.2}{13.6} \times \frac{36}{5} = 25$
$Z = 5$

AP EAMCET (18.09.2020) Shift-I

ATOMS

145563 In hydrogen spectrum, if the shortest wavelength in Balmer series is $λ$ , the shortest wavelength in Brackett series is

1

λ

2

λ / 2

3

4 λ

4

9 λ

Explanation:

C Shortest wavelength for electron transit from $\infty$ to $n$
$\frac{1}{λ} = R (\frac{1}{n^{2}})$
For Balmer series, $n = 2$
$\frac{1}{λ} = R (\frac{1}{2^{2}})$
$\frac{1}{λ} = \frac{R}{4} \Rightarrow R = \frac{4}{λ}$
For Brackett series, $n = 4$
$\frac{1}{λ (b)} = R (\frac{1}{n^{2}}) = \frac{R}{16}$
$\frac{1}{λ (b)} = \frac{R}{16}$
$\frac{16}{λ (b)} = R$
$\frac{16}{λ (b)} = \frac{4}{λ}$
$4 λ (b) = 16 λ$
$λ (b) = 4 λ$

TS- EAMCET-09.09.2020

ATOMS

145564 If the first line in the Lyman series has wavelength $λ$ , then the first line in Balmer series has the wavelength

1

\frac{27}{5} λ

2

\frac{32}{27} λ

3

\frac{28}{21} λ

4

\frac{15}{4} λ

Explanation:

A For hydrogen atom,
For first line in Lyman series
$n_{1} = 1, and n_{2} = 2$
$\frac{1}{λ_{L}} = R (\frac{1}{1^{2}} - \frac{1}{2^{2}})$
$\frac{1}{λ_{L}} = R (\frac{1}{1} - \frac{1}{4})$
$\frac{1}{λ_{L}} = \frac{3 R}{4}$
For first line in Balmer series $n_{1} = 2, and n_{2} = 3$
$\frac{1}{λ_{B}} = R (\frac{1}{2^{2}} - \frac{1}{3^{2}})$
$\frac{1}{λ_{B}} = R (\frac{1}{4} - \frac{1}{9})$
$\frac{1}{λ_{B}} = R (\frac{9 - 4}{36})$
$\frac{1}{λ_{B}} = \frac{5 R}{36}$
From equation (i) and equation (ii), we get-
$\frac{1}{λ_{B}} = \frac{5 R}{\frac{1}{36}}$
$\frac{λ_{L}}{4 R}$
$\frac{1 / λ_{B}}{1 / λ_{L}} = \frac{5 R}{36} \times \frac{4}{3 R}$
$\frac{λ_{L}}{λ_{B}} = \frac{5 R}{27 R}$
$λ_{B} = \frac{27}{5} λ_{l}$
$(λ_{L} = λ)$
$λ_{B} = \frac{27}{5} λ$

TS- EAMCET.14.09.2020

ATOMS

145559 The second line of Balmer series has wavelength $4861 \AA$ . The wavelength of the first line of Balmer series is

1

1216 \AA

2

6563 \AA

3

4340 \AA

4

4101 \AA

Explanation:

B For the first line Balmer series
$\frac{1}{λ} = R (\frac{1}{2^{2}} - \frac{1}{3^{2}}) = \frac{5 R}{36}$
For second line Balmer series.
$\frac{1}{4861} = R (\frac{1}{2^{2}} - \frac{1}{4^{2}}) = \frac{3 R}{16}$
divide equation (ii) by (i) then
$\frac{λ}{4861} = \frac{3 R}{16} \times \frac{36}{5 R}$
$λ = 4861 \times \frac{27}{20} = 6563 \AA$

AP EAMCET (18.09.2020) Shift-II

ATOMS

145561 If $λ_{1}$ and $λ_{2}$ are the wavelengths of the first members of the Lyman and Paschen series respectively, then $λ_{1} : λ_{2}$ is

1

1 : 3

2

1 : 30

3

7 : 50

4

7 : 108

Explanation:

D In the Lyman series, the first line is
$n_{1} = 1$ and $n_{2} = 2$
$\frac{1}{λ_{1}} = R [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$
$\frac{1}{λ_{1}} = R [1 - \frac{1}{2^{2}}]$
$\frac{1}{λ_{1}} = R [1 - \frac{1}{4}]$
$\frac{1}{λ_{1}} = \frac{3 R}{4}$
$λ_{1} = \frac{4}{3 R}$
Similarly in Paschen series the first line is
$n_{1} = 3 and n_{2} = 4$
$\frac{1}{λ_{2}} = R [\frac{1}{3^{2}} - \frac{1}{4^{2}}]$
$\frac{1}{λ_{2}} = R [\frac{1}{9} - \frac{1}{16}]$
$\frac{1}{λ_{2}} = \frac{7}{144} R$
$λ_{2} = \frac{144}{7 R}$
From (i) and (ii)
$\frac{λ_{1}}{λ_{2}} = \frac{4}{3 R} \times \frac{7 R}{144}$
$\frac{λ_{1}}{λ_{2}} = \frac{7}{108}$
$∴ λ_{1} : λ_{2} = 7 : 108$

MHT CET-2020

ATOMS

145562 A hydrogen like atom has one electron revolving round a stationary nucleus. If the energy required to excite the electron from the $2^{nd}$ orbital to $3^{rd}$ orbit is $47.2 eV$ , find the atomic number of the given atom.

1 3

2 4

3 5

4 6

Explanation:

C Given that, $Δ E = 47.2 eV, n_{1} = 2, n_{2} = 3$
We know that,
$Δ E = 13.6 Z^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$
$Δ E = 13.6 \times Z^{2} \times (\frac{1}{2^{2}} - \frac{1}{3^{2}})$
$47.2 = 13.6 Z^{2} \times \frac{5}{36}$
$Z^{2} = \frac{47.2}{13.6} \times \frac{36}{5} = 25$
$Z = 5$

AP EAMCET (18.09.2020) Shift-I

ATOMS

145563 In hydrogen spectrum, if the shortest wavelength in Balmer series is $λ$ , the shortest wavelength in Brackett series is

1

λ

2

λ / 2

3

4 λ

4

9 λ

Explanation:

C Shortest wavelength for electron transit from $\infty$ to $n$
$\frac{1}{λ} = R (\frac{1}{n^{2}})$
For Balmer series, $n = 2$
$\frac{1}{λ} = R (\frac{1}{2^{2}})$
$\frac{1}{λ} = \frac{R}{4} \Rightarrow R = \frac{4}{λ}$
For Brackett series, $n = 4$
$\frac{1}{λ (b)} = R (\frac{1}{n^{2}}) = \frac{R}{16}$
$\frac{1}{λ (b)} = \frac{R}{16}$
$\frac{16}{λ (b)} = R$
$\frac{16}{λ (b)} = \frac{4}{λ}$
$4 λ (b) = 16 λ$
$λ (b) = 4 λ$

TS- EAMCET-09.09.2020

ATOMS

145564 If the first line in the Lyman series has wavelength $λ$ , then the first line in Balmer series has the wavelength

1

\frac{27}{5} λ

2

\frac{32}{27} λ

3

\frac{28}{21} λ

4

\frac{15}{4} λ

Explanation:

A For hydrogen atom,
For first line in Lyman series
$n_{1} = 1, and n_{2} = 2$
$\frac{1}{λ_{L}} = R (\frac{1}{1^{2}} - \frac{1}{2^{2}})$
$\frac{1}{λ_{L}} = R (\frac{1}{1} - \frac{1}{4})$
$\frac{1}{λ_{L}} = \frac{3 R}{4}$
For first line in Balmer series $n_{1} = 2, and n_{2} = 3$
$\frac{1}{λ_{B}} = R (\frac{1}{2^{2}} - \frac{1}{3^{2}})$
$\frac{1}{λ_{B}} = R (\frac{1}{4} - \frac{1}{9})$
$\frac{1}{λ_{B}} = R (\frac{9 - 4}{36})$
$\frac{1}{λ_{B}} = \frac{5 R}{36}$
From equation (i) and equation (ii), we get-
$\frac{1}{λ_{B}} = \frac{5 R}{\frac{1}{36}}$
$\frac{λ_{L}}{4 R}$
$\frac{1 / λ_{B}}{1 / λ_{L}} = \frac{5 R}{36} \times \frac{4}{3 R}$
$\frac{λ_{L}}{λ_{B}} = \frac{5 R}{27 R}$
$λ_{B} = \frac{27}{5} λ_{l}$
$(λ_{L} = λ)$
$λ_{B} = \frac{27}{5} λ$

TS- EAMCET.14.09.2020

ATOMS

145559 The second line of Balmer series has wavelength $4861 \AA$ . The wavelength of the first line of Balmer series is

1

1216 \AA

2

6563 \AA

3

4340 \AA

4

4101 \AA

Explanation:

B For the first line Balmer series
$\frac{1}{λ} = R (\frac{1}{2^{2}} - \frac{1}{3^{2}}) = \frac{5 R}{36}$
For second line Balmer series.
$\frac{1}{4861} = R (\frac{1}{2^{2}} - \frac{1}{4^{2}}) = \frac{3 R}{16}$
divide equation (ii) by (i) then
$\frac{λ}{4861} = \frac{3 R}{16} \times \frac{36}{5 R}$
$λ = 4861 \times \frac{27}{20} = 6563 \AA$

AP EAMCET (18.09.2020) Shift-II

ATOMS

145561 If $λ_{1}$ and $λ_{2}$ are the wavelengths of the first members of the Lyman and Paschen series respectively, then $λ_{1} : λ_{2}$ is

1

1 : 3

2

1 : 30

3

7 : 50

4

7 : 108

Explanation:

D In the Lyman series, the first line is
$n_{1} = 1$ and $n_{2} = 2$
$\frac{1}{λ_{1}} = R [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$
$\frac{1}{λ_{1}} = R [1 - \frac{1}{2^{2}}]$
$\frac{1}{λ_{1}} = R [1 - \frac{1}{4}]$
$\frac{1}{λ_{1}} = \frac{3 R}{4}$
$λ_{1} = \frac{4}{3 R}$
Similarly in Paschen series the first line is
$n_{1} = 3 and n_{2} = 4$
$\frac{1}{λ_{2}} = R [\frac{1}{3^{2}} - \frac{1}{4^{2}}]$
$\frac{1}{λ_{2}} = R [\frac{1}{9} - \frac{1}{16}]$
$\frac{1}{λ_{2}} = \frac{7}{144} R$
$λ_{2} = \frac{144}{7 R}$
From (i) and (ii)
$\frac{λ_{1}}{λ_{2}} = \frac{4}{3 R} \times \frac{7 R}{144}$
$\frac{λ_{1}}{λ_{2}} = \frac{7}{108}$
$∴ λ_{1} : λ_{2} = 7 : 108$

MHT CET-2020

ATOMS

145562 A hydrogen like atom has one electron revolving round a stationary nucleus. If the energy required to excite the electron from the $2^{nd}$ orbital to $3^{rd}$ orbit is $47.2 eV$ , find the atomic number of the given atom.

1 3

2 4

3 5

4 6

Explanation:

C Given that, $Δ E = 47.2 eV, n_{1} = 2, n_{2} = 3$
We know that,
$Δ E = 13.6 Z^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$
$Δ E = 13.6 \times Z^{2} \times (\frac{1}{2^{2}} - \frac{1}{3^{2}})$
$47.2 = 13.6 Z^{2} \times \frac{5}{36}$
$Z^{2} = \frac{47.2}{13.6} \times \frac{36}{5} = 25$
$Z = 5$

AP EAMCET (18.09.2020) Shift-I

ATOMS

145563 In hydrogen spectrum, if the shortest wavelength in Balmer series is $λ$ , the shortest wavelength in Brackett series is

1

λ

2

λ / 2

3

4 λ

4

9 λ

Explanation:

C Shortest wavelength for electron transit from $\infty$ to $n$
$\frac{1}{λ} = R (\frac{1}{n^{2}})$
For Balmer series, $n = 2$
$\frac{1}{λ} = R (\frac{1}{2^{2}})$
$\frac{1}{λ} = \frac{R}{4} \Rightarrow R = \frac{4}{λ}$
For Brackett series, $n = 4$
$\frac{1}{λ (b)} = R (\frac{1}{n^{2}}) = \frac{R}{16}$
$\frac{1}{λ (b)} = \frac{R}{16}$
$\frac{16}{λ (b)} = R$
$\frac{16}{λ (b)} = \frac{4}{λ}$
$4 λ (b) = 16 λ$
$λ (b) = 4 λ$

TS- EAMCET-09.09.2020

ATOMS

145564 If the first line in the Lyman series has wavelength $λ$ , then the first line in Balmer series has the wavelength

1

\frac{27}{5} λ

2

\frac{32}{27} λ

3

\frac{28}{21} λ

4

\frac{15}{4} λ

Explanation:

A For hydrogen atom,
For first line in Lyman series
$n_{1} = 1, and n_{2} = 2$
$\frac{1}{λ_{L}} = R (\frac{1}{1^{2}} - \frac{1}{2^{2}})$
$\frac{1}{λ_{L}} = R (\frac{1}{1} - \frac{1}{4})$
$\frac{1}{λ_{L}} = \frac{3 R}{4}$
For first line in Balmer series $n_{1} = 2, and n_{2} = 3$
$\frac{1}{λ_{B}} = R (\frac{1}{2^{2}} - \frac{1}{3^{2}})$
$\frac{1}{λ_{B}} = R (\frac{1}{4} - \frac{1}{9})$
$\frac{1}{λ_{B}} = R (\frac{9 - 4}{36})$
$\frac{1}{λ_{B}} = \frac{5 R}{36}$
From equation (i) and equation (ii), we get-
$\frac{1}{λ_{B}} = \frac{5 R}{\frac{1}{36}}$
$\frac{λ_{L}}{4 R}$
$\frac{1 / λ_{B}}{1 / λ_{L}} = \frac{5 R}{36} \times \frac{4}{3 R}$
$\frac{λ_{L}}{λ_{B}} = \frac{5 R}{27 R}$
$λ_{B} = \frac{27}{5} λ_{l}$
$(λ_{L} = λ)$
$λ_{B} = \frac{27}{5} λ$

TS- EAMCET.14.09.2020

ATOMS

145559 The second line of Balmer series has wavelength $4861 \AA$ . The wavelength of the first line of Balmer series is

1

1216 \AA

2

6563 \AA

3

4340 \AA

4

4101 \AA

Explanation:

B For the first line Balmer series
$\frac{1}{λ} = R (\frac{1}{2^{2}} - \frac{1}{3^{2}}) = \frac{5 R}{36}$
For second line Balmer series.
$\frac{1}{4861} = R (\frac{1}{2^{2}} - \frac{1}{4^{2}}) = \frac{3 R}{16}$
divide equation (ii) by (i) then
$\frac{λ}{4861} = \frac{3 R}{16} \times \frac{36}{5 R}$
$λ = 4861 \times \frac{27}{20} = 6563 \AA$

AP EAMCET (18.09.2020) Shift-II

ATOMS

145561 If $λ_{1}$ and $λ_{2}$ are the wavelengths of the first members of the Lyman and Paschen series respectively, then $λ_{1} : λ_{2}$ is

1

1 : 3

2

1 : 30

3

7 : 50

4

7 : 108

Explanation:

D In the Lyman series, the first line is
$n_{1} = 1$ and $n_{2} = 2$
$\frac{1}{λ_{1}} = R [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$
$\frac{1}{λ_{1}} = R [1 - \frac{1}{2^{2}}]$
$\frac{1}{λ_{1}} = R [1 - \frac{1}{4}]$
$\frac{1}{λ_{1}} = \frac{3 R}{4}$
$λ_{1} = \frac{4}{3 R}$
Similarly in Paschen series the first line is
$n_{1} = 3 and n_{2} = 4$
$\frac{1}{λ_{2}} = R [\frac{1}{3^{2}} - \frac{1}{4^{2}}]$
$\frac{1}{λ_{2}} = R [\frac{1}{9} - \frac{1}{16}]$
$\frac{1}{λ_{2}} = \frac{7}{144} R$
$λ_{2} = \frac{144}{7 R}$
From (i) and (ii)
$\frac{λ_{1}}{λ_{2}} = \frac{4}{3 R} \times \frac{7 R}{144}$
$\frac{λ_{1}}{λ_{2}} = \frac{7}{108}$
$∴ λ_{1} : λ_{2} = 7 : 108$

MHT CET-2020

ATOMS

145562 A hydrogen like atom has one electron revolving round a stationary nucleus. If the energy required to excite the electron from the $2^{nd}$ orbital to $3^{rd}$ orbit is $47.2 eV$ , find the atomic number of the given atom.

1 3

2 4

3 5

4 6

Explanation:

C Given that, $Δ E = 47.2 eV, n_{1} = 2, n_{2} = 3$
We know that,
$Δ E = 13.6 Z^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$
$Δ E = 13.6 \times Z^{2} \times (\frac{1}{2^{2}} - \frac{1}{3^{2}})$
$47.2 = 13.6 Z^{2} \times \frac{5}{36}$
$Z^{2} = \frac{47.2}{13.6} \times \frac{36}{5} = 25$
$Z = 5$

AP EAMCET (18.09.2020) Shift-I

ATOMS

145563 In hydrogen spectrum, if the shortest wavelength in Balmer series is $λ$ , the shortest wavelength in Brackett series is

1

λ

2

λ / 2

3

4 λ

4

9 λ

Explanation:

C Shortest wavelength for electron transit from $\infty$ to $n$
$\frac{1}{λ} = R (\frac{1}{n^{2}})$
For Balmer series, $n = 2$
$\frac{1}{λ} = R (\frac{1}{2^{2}})$
$\frac{1}{λ} = \frac{R}{4} \Rightarrow R = \frac{4}{λ}$
For Brackett series, $n = 4$
$\frac{1}{λ (b)} = R (\frac{1}{n^{2}}) = \frac{R}{16}$
$\frac{1}{λ (b)} = \frac{R}{16}$
$\frac{16}{λ (b)} = R$
$\frac{16}{λ (b)} = \frac{4}{λ}$
$4 λ (b) = 16 λ$
$λ (b) = 4 λ$

TS- EAMCET-09.09.2020

ATOMS

145564 If the first line in the Lyman series has wavelength $λ$ , then the first line in Balmer series has the wavelength

1

\frac{27}{5} λ

2

\frac{32}{27} λ

3

\frac{28}{21} λ

4

\frac{15}{4} λ

Explanation:

A For hydrogen atom,
For first line in Lyman series
$n_{1} = 1, and n_{2} = 2$
$\frac{1}{λ_{L}} = R (\frac{1}{1^{2}} - \frac{1}{2^{2}})$
$\frac{1}{λ_{L}} = R (\frac{1}{1} - \frac{1}{4})$
$\frac{1}{λ_{L}} = \frac{3 R}{4}$
For first line in Balmer series $n_{1} = 2, and n_{2} = 3$
$\frac{1}{λ_{B}} = R (\frac{1}{2^{2}} - \frac{1}{3^{2}})$
$\frac{1}{λ_{B}} = R (\frac{1}{4} - \frac{1}{9})$
$\frac{1}{λ_{B}} = R (\frac{9 - 4}{36})$
$\frac{1}{λ_{B}} = \frac{5 R}{36}$
From equation (i) and equation (ii), we get-
$\frac{1}{λ_{B}} = \frac{5 R}{\frac{1}{36}}$
$\frac{λ_{L}}{4 R}$
$\frac{1 / λ_{B}}{1 / λ_{L}} = \frac{5 R}{36} \times \frac{4}{3 R}$
$\frac{λ_{L}}{λ_{B}} = \frac{5 R}{27 R}$
$λ_{B} = \frac{27}{5} λ_{l}$
$(λ_{L} = λ)$
$λ_{B} = \frac{27}{5} λ$

TS- EAMCET.14.09.2020