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 $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=\frac{5 \mathrm{R}}{36}$ For second line Balmer series. $\frac{1}{4861}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{4^{2}}\right)=\frac{3 \mathrm{R}}{16}$ divide equation (ii) by (i) then $\frac{\lambda}{4861}=\frac{3 \mathrm{R}}{16} \times \frac{36}{5 \mathrm{R}}$ $\lambda=4861 \times \frac{27}{20}=6563 \AA$
AP EAMCET (18.09.2020) Shift-II
ATOMS
145561
If $\lambda_{1}$ and $\lambda_{2}$ are the wavelengths of the first members of the Lyman and Paschen series respectively, then $\lambda_{1}: \lambda_{2}$ is
1 $1: 3$
2 $1: 30$
3 $7: 50$
4 $7: 108$
Explanation:
D In the Lyman series, the first line is $\mathrm{n}_{1}=1$ and $\mathrm{n}_{2}=2$ $\frac{1}{\lambda_{1}}=\mathrm{R}\left[\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right]$ $\frac{1}{\lambda_{1}}=\mathrm{R}\left[1-\frac{1}{2^{2}}\right]$ $\frac{1}{\lambda_{1}}=\mathrm{R}\left[1-\frac{1}{4}\right]$ $\frac{1}{\lambda_{1}}=\frac{3 \mathrm{R}}{4}$ $\lambda_{1}=\frac{4}{3 \mathrm{R}}$ Similarly in Paschen series the first line is $\mathrm{n}_{1}=3 \text { and } \mathrm{n}_{2}=4$ $\frac{1}{\lambda_{2}}=\mathrm{R}\left[\frac{1}{3^{2}}-\frac{1}{4^{2}}\right]$ $\frac{1}{\lambda_{2}}=\mathrm{R}\left[\frac{1}{9}-\frac{1}{16}\right]$ $\frac{1}{\lambda_{2}}=\frac{7}{144} \mathrm{R}$ $\lambda_{2}=\frac{144}{7 \mathrm{R}}$ From (i) and (ii) $\frac{\lambda_{1}}{\lambda_{2}}=\frac{4}{3 \mathrm{R}} \times \frac{7 \mathrm{R}}{144}$ $\frac{\lambda_{1}}{\lambda_{2}}=\frac{7}{108}$ $\therefore \quad \lambda_{1}: \lambda_{2}=7: 108$
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 $2^{\text {nd }}$ orbital to $3^{\text {rd }}$ orbit is $47.2 \mathrm{eV}$, find the atomic number of the given atom.
1 3
2 4
3 5
4 6
Explanation:
C Given that, $\Delta \mathrm{E}=47.2 \mathrm{eV}, \mathrm{n}_{1}=2, \mathrm{n}_{2}=3$ We know that, $\Delta \mathrm{E}=13.6 \mathrm{Z}^{2}\left(\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right)$ $\Delta \mathrm{E}=13.6 \times \mathrm{Z}^{2} \times\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)$ $47.2=13.6 \mathrm{Z}^{2} \times \frac{5}{36}$ $\mathrm{Z}^{2}=\frac{47.2}{13.6} \times \frac{36}{5}=25$ $\mathrm{Z}=5$
AP EAMCET (18.09.2020) Shift-I
ATOMS
145563
In hydrogen spectrum, if the shortest wavelength in Balmer series is $\lambda$, the shortest wavelength in Brackett series is
1 $\lambda$
2 $\lambda / 2$
3 $4 \lambda$
4 $9 \lambda$
Explanation:
C Shortest wavelength for electron transit from $\infty$ to $\mathrm{n}$ $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{\mathrm{n}^{2}}\right)$ For Balmer series, $\mathrm{n}=2$ $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{2^{2}}\right)$ $\frac{1}{\lambda}=\frac{\mathrm{R}}{4} \Rightarrow \mathrm{R}=\frac{4}{\lambda}$ For Brackett series, $n=4$ $\frac{1}{\lambda(\mathrm{b})}=\mathrm{R}\left(\frac{1}{\mathrm{n}^{2}}\right)=\frac{\mathrm{R}}{16}$ $\frac{1}{\lambda(\mathrm{b})}=\frac{\mathrm{R}}{16}$ \(\frac{16}{\lambda(\mathrm{b})}=\mathrm{R}\) \(\frac{16}{\lambda(\mathrm{b})}=\frac{4}{\lambda}\) \(4 \lambda(\mathrm{b})=16 \lambda\) \(\lambda(\mathrm{b})=4 \lambda\)
TS- EAMCET-09.09.2020
ATOMS
145564
If the first line in the Lyman series has wavelength $\lambda$, then the first line in Balmer series has the wavelength
1 $\frac{27}{5} \lambda$
2 $\frac{32}{27} \lambda$
3 $\frac{28}{21} \lambda$
4 $\frac{15}{4} \lambda$
Explanation:
A For hydrogen atom, For first line in Lyman series $\mathrm{n}_{1}=1, \text { and } \mathrm{n}_{2}=2$ $\frac{1}{\lambda_{\mathrm{L}}}=\mathrm{R}\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)$ $\frac{1}{\lambda_{\mathrm{L}}}=\mathrm{R}\left(\frac{1}{1}-\frac{1}{4}\right)$ $\frac{1}{\lambda_{\mathrm{L}}}=\frac{3 \mathrm{R}}{4}$ For first line in Balmer series$\mathrm{n}_{1}=2 \text {, and } \mathrm{n}_{2}=3$ $\frac{1}{\lambda_{\mathrm{B}}}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)$ $\frac{1}{\lambda_{\mathrm{B}}}=\mathrm{R}\left(\frac{1}{4}-\frac{1}{9}\right)$ $\frac{1}{\lambda_{\mathrm{B}}}=\mathrm{R}\left(\frac{9-4}{36}\right)$ $\frac{1}{\lambda_{\mathrm{B}}}=\frac{5 \mathrm{R}}{36}$ From equation (i) and equation (ii), we get- $\frac{1}{\lambda_{\mathrm{B}}}=\frac{5 \mathrm{R}}{\frac{1}{36}}$ $\frac{\lambda_{\mathrm{L}}}{4 \mathrm{R}}$ $\frac{1 / \lambda_{\mathrm{B}}}{1 / \lambda_{\mathrm{L}}}=\frac{5 \mathrm{R}}{36} \times \frac{4}{3 \mathrm{R}}$ $\frac{\lambda_{L}}{\lambda_{\mathrm{B}}}=\frac{5 \mathrm{R}}{27 \mathrm{R}}$ $\lambda_{\mathrm{B}}=\frac{27}{5} \lambda_{l}$ $\left(\lambda_{\mathrm{L}}=\lambda\right)$ $\lambda_{\mathrm{B}}=\frac{27}{5} \lambda$
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 $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=\frac{5 \mathrm{R}}{36}$ For second line Balmer series. $\frac{1}{4861}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{4^{2}}\right)=\frac{3 \mathrm{R}}{16}$ divide equation (ii) by (i) then $\frac{\lambda}{4861}=\frac{3 \mathrm{R}}{16} \times \frac{36}{5 \mathrm{R}}$ $\lambda=4861 \times \frac{27}{20}=6563 \AA$
AP EAMCET (18.09.2020) Shift-II
ATOMS
145561
If $\lambda_{1}$ and $\lambda_{2}$ are the wavelengths of the first members of the Lyman and Paschen series respectively, then $\lambda_{1}: \lambda_{2}$ is
1 $1: 3$
2 $1: 30$
3 $7: 50$
4 $7: 108$
Explanation:
D In the Lyman series, the first line is $\mathrm{n}_{1}=1$ and $\mathrm{n}_{2}=2$ $\frac{1}{\lambda_{1}}=\mathrm{R}\left[\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right]$ $\frac{1}{\lambda_{1}}=\mathrm{R}\left[1-\frac{1}{2^{2}}\right]$ $\frac{1}{\lambda_{1}}=\mathrm{R}\left[1-\frac{1}{4}\right]$ $\frac{1}{\lambda_{1}}=\frac{3 \mathrm{R}}{4}$ $\lambda_{1}=\frac{4}{3 \mathrm{R}}$ Similarly in Paschen series the first line is $\mathrm{n}_{1}=3 \text { and } \mathrm{n}_{2}=4$ $\frac{1}{\lambda_{2}}=\mathrm{R}\left[\frac{1}{3^{2}}-\frac{1}{4^{2}}\right]$ $\frac{1}{\lambda_{2}}=\mathrm{R}\left[\frac{1}{9}-\frac{1}{16}\right]$ $\frac{1}{\lambda_{2}}=\frac{7}{144} \mathrm{R}$ $\lambda_{2}=\frac{144}{7 \mathrm{R}}$ From (i) and (ii) $\frac{\lambda_{1}}{\lambda_{2}}=\frac{4}{3 \mathrm{R}} \times \frac{7 \mathrm{R}}{144}$ $\frac{\lambda_{1}}{\lambda_{2}}=\frac{7}{108}$ $\therefore \quad \lambda_{1}: \lambda_{2}=7: 108$
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 $2^{\text {nd }}$ orbital to $3^{\text {rd }}$ orbit is $47.2 \mathrm{eV}$, find the atomic number of the given atom.
1 3
2 4
3 5
4 6
Explanation:
C Given that, $\Delta \mathrm{E}=47.2 \mathrm{eV}, \mathrm{n}_{1}=2, \mathrm{n}_{2}=3$ We know that, $\Delta \mathrm{E}=13.6 \mathrm{Z}^{2}\left(\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right)$ $\Delta \mathrm{E}=13.6 \times \mathrm{Z}^{2} \times\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)$ $47.2=13.6 \mathrm{Z}^{2} \times \frac{5}{36}$ $\mathrm{Z}^{2}=\frac{47.2}{13.6} \times \frac{36}{5}=25$ $\mathrm{Z}=5$
AP EAMCET (18.09.2020) Shift-I
ATOMS
145563
In hydrogen spectrum, if the shortest wavelength in Balmer series is $\lambda$, the shortest wavelength in Brackett series is
1 $\lambda$
2 $\lambda / 2$
3 $4 \lambda$
4 $9 \lambda$
Explanation:
C Shortest wavelength for electron transit from $\infty$ to $\mathrm{n}$ $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{\mathrm{n}^{2}}\right)$ For Balmer series, $\mathrm{n}=2$ $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{2^{2}}\right)$ $\frac{1}{\lambda}=\frac{\mathrm{R}}{4} \Rightarrow \mathrm{R}=\frac{4}{\lambda}$ For Brackett series, $n=4$ $\frac{1}{\lambda(\mathrm{b})}=\mathrm{R}\left(\frac{1}{\mathrm{n}^{2}}\right)=\frac{\mathrm{R}}{16}$ $\frac{1}{\lambda(\mathrm{b})}=\frac{\mathrm{R}}{16}$ \(\frac{16}{\lambda(\mathrm{b})}=\mathrm{R}\) \(\frac{16}{\lambda(\mathrm{b})}=\frac{4}{\lambda}\) \(4 \lambda(\mathrm{b})=16 \lambda\) \(\lambda(\mathrm{b})=4 \lambda\)
TS- EAMCET-09.09.2020
ATOMS
145564
If the first line in the Lyman series has wavelength $\lambda$, then the first line in Balmer series has the wavelength
1 $\frac{27}{5} \lambda$
2 $\frac{32}{27} \lambda$
3 $\frac{28}{21} \lambda$
4 $\frac{15}{4} \lambda$
Explanation:
A For hydrogen atom, For first line in Lyman series $\mathrm{n}_{1}=1, \text { and } \mathrm{n}_{2}=2$ $\frac{1}{\lambda_{\mathrm{L}}}=\mathrm{R}\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)$ $\frac{1}{\lambda_{\mathrm{L}}}=\mathrm{R}\left(\frac{1}{1}-\frac{1}{4}\right)$ $\frac{1}{\lambda_{\mathrm{L}}}=\frac{3 \mathrm{R}}{4}$ For first line in Balmer series$\mathrm{n}_{1}=2 \text {, and } \mathrm{n}_{2}=3$ $\frac{1}{\lambda_{\mathrm{B}}}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)$ $\frac{1}{\lambda_{\mathrm{B}}}=\mathrm{R}\left(\frac{1}{4}-\frac{1}{9}\right)$ $\frac{1}{\lambda_{\mathrm{B}}}=\mathrm{R}\left(\frac{9-4}{36}\right)$ $\frac{1}{\lambda_{\mathrm{B}}}=\frac{5 \mathrm{R}}{36}$ From equation (i) and equation (ii), we get- $\frac{1}{\lambda_{\mathrm{B}}}=\frac{5 \mathrm{R}}{\frac{1}{36}}$ $\frac{\lambda_{\mathrm{L}}}{4 \mathrm{R}}$ $\frac{1 / \lambda_{\mathrm{B}}}{1 / \lambda_{\mathrm{L}}}=\frac{5 \mathrm{R}}{36} \times \frac{4}{3 \mathrm{R}}$ $\frac{\lambda_{L}}{\lambda_{\mathrm{B}}}=\frac{5 \mathrm{R}}{27 \mathrm{R}}$ $\lambda_{\mathrm{B}}=\frac{27}{5} \lambda_{l}$ $\left(\lambda_{\mathrm{L}}=\lambda\right)$ $\lambda_{\mathrm{B}}=\frac{27}{5} \lambda$
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 $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=\frac{5 \mathrm{R}}{36}$ For second line Balmer series. $\frac{1}{4861}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{4^{2}}\right)=\frac{3 \mathrm{R}}{16}$ divide equation (ii) by (i) then $\frac{\lambda}{4861}=\frac{3 \mathrm{R}}{16} \times \frac{36}{5 \mathrm{R}}$ $\lambda=4861 \times \frac{27}{20}=6563 \AA$
AP EAMCET (18.09.2020) Shift-II
ATOMS
145561
If $\lambda_{1}$ and $\lambda_{2}$ are the wavelengths of the first members of the Lyman and Paschen series respectively, then $\lambda_{1}: \lambda_{2}$ is
1 $1: 3$
2 $1: 30$
3 $7: 50$
4 $7: 108$
Explanation:
D In the Lyman series, the first line is $\mathrm{n}_{1}=1$ and $\mathrm{n}_{2}=2$ $\frac{1}{\lambda_{1}}=\mathrm{R}\left[\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right]$ $\frac{1}{\lambda_{1}}=\mathrm{R}\left[1-\frac{1}{2^{2}}\right]$ $\frac{1}{\lambda_{1}}=\mathrm{R}\left[1-\frac{1}{4}\right]$ $\frac{1}{\lambda_{1}}=\frac{3 \mathrm{R}}{4}$ $\lambda_{1}=\frac{4}{3 \mathrm{R}}$ Similarly in Paschen series the first line is $\mathrm{n}_{1}=3 \text { and } \mathrm{n}_{2}=4$ $\frac{1}{\lambda_{2}}=\mathrm{R}\left[\frac{1}{3^{2}}-\frac{1}{4^{2}}\right]$ $\frac{1}{\lambda_{2}}=\mathrm{R}\left[\frac{1}{9}-\frac{1}{16}\right]$ $\frac{1}{\lambda_{2}}=\frac{7}{144} \mathrm{R}$ $\lambda_{2}=\frac{144}{7 \mathrm{R}}$ From (i) and (ii) $\frac{\lambda_{1}}{\lambda_{2}}=\frac{4}{3 \mathrm{R}} \times \frac{7 \mathrm{R}}{144}$ $\frac{\lambda_{1}}{\lambda_{2}}=\frac{7}{108}$ $\therefore \quad \lambda_{1}: \lambda_{2}=7: 108$
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 $2^{\text {nd }}$ orbital to $3^{\text {rd }}$ orbit is $47.2 \mathrm{eV}$, find the atomic number of the given atom.
1 3
2 4
3 5
4 6
Explanation:
C Given that, $\Delta \mathrm{E}=47.2 \mathrm{eV}, \mathrm{n}_{1}=2, \mathrm{n}_{2}=3$ We know that, $\Delta \mathrm{E}=13.6 \mathrm{Z}^{2}\left(\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right)$ $\Delta \mathrm{E}=13.6 \times \mathrm{Z}^{2} \times\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)$ $47.2=13.6 \mathrm{Z}^{2} \times \frac{5}{36}$ $\mathrm{Z}^{2}=\frac{47.2}{13.6} \times \frac{36}{5}=25$ $\mathrm{Z}=5$
AP EAMCET (18.09.2020) Shift-I
ATOMS
145563
In hydrogen spectrum, if the shortest wavelength in Balmer series is $\lambda$, the shortest wavelength in Brackett series is
1 $\lambda$
2 $\lambda / 2$
3 $4 \lambda$
4 $9 \lambda$
Explanation:
C Shortest wavelength for electron transit from $\infty$ to $\mathrm{n}$ $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{\mathrm{n}^{2}}\right)$ For Balmer series, $\mathrm{n}=2$ $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{2^{2}}\right)$ $\frac{1}{\lambda}=\frac{\mathrm{R}}{4} \Rightarrow \mathrm{R}=\frac{4}{\lambda}$ For Brackett series, $n=4$ $\frac{1}{\lambda(\mathrm{b})}=\mathrm{R}\left(\frac{1}{\mathrm{n}^{2}}\right)=\frac{\mathrm{R}}{16}$ $\frac{1}{\lambda(\mathrm{b})}=\frac{\mathrm{R}}{16}$ \(\frac{16}{\lambda(\mathrm{b})}=\mathrm{R}\) \(\frac{16}{\lambda(\mathrm{b})}=\frac{4}{\lambda}\) \(4 \lambda(\mathrm{b})=16 \lambda\) \(\lambda(\mathrm{b})=4 \lambda\)
TS- EAMCET-09.09.2020
ATOMS
145564
If the first line in the Lyman series has wavelength $\lambda$, then the first line in Balmer series has the wavelength
1 $\frac{27}{5} \lambda$
2 $\frac{32}{27} \lambda$
3 $\frac{28}{21} \lambda$
4 $\frac{15}{4} \lambda$
Explanation:
A For hydrogen atom, For first line in Lyman series $\mathrm{n}_{1}=1, \text { and } \mathrm{n}_{2}=2$ $\frac{1}{\lambda_{\mathrm{L}}}=\mathrm{R}\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)$ $\frac{1}{\lambda_{\mathrm{L}}}=\mathrm{R}\left(\frac{1}{1}-\frac{1}{4}\right)$ $\frac{1}{\lambda_{\mathrm{L}}}=\frac{3 \mathrm{R}}{4}$ For first line in Balmer series$\mathrm{n}_{1}=2 \text {, and } \mathrm{n}_{2}=3$ $\frac{1}{\lambda_{\mathrm{B}}}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)$ $\frac{1}{\lambda_{\mathrm{B}}}=\mathrm{R}\left(\frac{1}{4}-\frac{1}{9}\right)$ $\frac{1}{\lambda_{\mathrm{B}}}=\mathrm{R}\left(\frac{9-4}{36}\right)$ $\frac{1}{\lambda_{\mathrm{B}}}=\frac{5 \mathrm{R}}{36}$ From equation (i) and equation (ii), we get- $\frac{1}{\lambda_{\mathrm{B}}}=\frac{5 \mathrm{R}}{\frac{1}{36}}$ $\frac{\lambda_{\mathrm{L}}}{4 \mathrm{R}}$ $\frac{1 / \lambda_{\mathrm{B}}}{1 / \lambda_{\mathrm{L}}}=\frac{5 \mathrm{R}}{36} \times \frac{4}{3 \mathrm{R}}$ $\frac{\lambda_{L}}{\lambda_{\mathrm{B}}}=\frac{5 \mathrm{R}}{27 \mathrm{R}}$ $\lambda_{\mathrm{B}}=\frac{27}{5} \lambda_{l}$ $\left(\lambda_{\mathrm{L}}=\lambda\right)$ $\lambda_{\mathrm{B}}=\frac{27}{5} \lambda$
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 $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=\frac{5 \mathrm{R}}{36}$ For second line Balmer series. $\frac{1}{4861}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{4^{2}}\right)=\frac{3 \mathrm{R}}{16}$ divide equation (ii) by (i) then $\frac{\lambda}{4861}=\frac{3 \mathrm{R}}{16} \times \frac{36}{5 \mathrm{R}}$ $\lambda=4861 \times \frac{27}{20}=6563 \AA$
AP EAMCET (18.09.2020) Shift-II
ATOMS
145561
If $\lambda_{1}$ and $\lambda_{2}$ are the wavelengths of the first members of the Lyman and Paschen series respectively, then $\lambda_{1}: \lambda_{2}$ is
1 $1: 3$
2 $1: 30$
3 $7: 50$
4 $7: 108$
Explanation:
D In the Lyman series, the first line is $\mathrm{n}_{1}=1$ and $\mathrm{n}_{2}=2$ $\frac{1}{\lambda_{1}}=\mathrm{R}\left[\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right]$ $\frac{1}{\lambda_{1}}=\mathrm{R}\left[1-\frac{1}{2^{2}}\right]$ $\frac{1}{\lambda_{1}}=\mathrm{R}\left[1-\frac{1}{4}\right]$ $\frac{1}{\lambda_{1}}=\frac{3 \mathrm{R}}{4}$ $\lambda_{1}=\frac{4}{3 \mathrm{R}}$ Similarly in Paschen series the first line is $\mathrm{n}_{1}=3 \text { and } \mathrm{n}_{2}=4$ $\frac{1}{\lambda_{2}}=\mathrm{R}\left[\frac{1}{3^{2}}-\frac{1}{4^{2}}\right]$ $\frac{1}{\lambda_{2}}=\mathrm{R}\left[\frac{1}{9}-\frac{1}{16}\right]$ $\frac{1}{\lambda_{2}}=\frac{7}{144} \mathrm{R}$ $\lambda_{2}=\frac{144}{7 \mathrm{R}}$ From (i) and (ii) $\frac{\lambda_{1}}{\lambda_{2}}=\frac{4}{3 \mathrm{R}} \times \frac{7 \mathrm{R}}{144}$ $\frac{\lambda_{1}}{\lambda_{2}}=\frac{7}{108}$ $\therefore \quad \lambda_{1}: \lambda_{2}=7: 108$
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 $2^{\text {nd }}$ orbital to $3^{\text {rd }}$ orbit is $47.2 \mathrm{eV}$, find the atomic number of the given atom.
1 3
2 4
3 5
4 6
Explanation:
C Given that, $\Delta \mathrm{E}=47.2 \mathrm{eV}, \mathrm{n}_{1}=2, \mathrm{n}_{2}=3$ We know that, $\Delta \mathrm{E}=13.6 \mathrm{Z}^{2}\left(\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right)$ $\Delta \mathrm{E}=13.6 \times \mathrm{Z}^{2} \times\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)$ $47.2=13.6 \mathrm{Z}^{2} \times \frac{5}{36}$ $\mathrm{Z}^{2}=\frac{47.2}{13.6} \times \frac{36}{5}=25$ $\mathrm{Z}=5$
AP EAMCET (18.09.2020) Shift-I
ATOMS
145563
In hydrogen spectrum, if the shortest wavelength in Balmer series is $\lambda$, the shortest wavelength in Brackett series is
1 $\lambda$
2 $\lambda / 2$
3 $4 \lambda$
4 $9 \lambda$
Explanation:
C Shortest wavelength for electron transit from $\infty$ to $\mathrm{n}$ $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{\mathrm{n}^{2}}\right)$ For Balmer series, $\mathrm{n}=2$ $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{2^{2}}\right)$ $\frac{1}{\lambda}=\frac{\mathrm{R}}{4} \Rightarrow \mathrm{R}=\frac{4}{\lambda}$ For Brackett series, $n=4$ $\frac{1}{\lambda(\mathrm{b})}=\mathrm{R}\left(\frac{1}{\mathrm{n}^{2}}\right)=\frac{\mathrm{R}}{16}$ $\frac{1}{\lambda(\mathrm{b})}=\frac{\mathrm{R}}{16}$ \(\frac{16}{\lambda(\mathrm{b})}=\mathrm{R}\) \(\frac{16}{\lambda(\mathrm{b})}=\frac{4}{\lambda}\) \(4 \lambda(\mathrm{b})=16 \lambda\) \(\lambda(\mathrm{b})=4 \lambda\)
TS- EAMCET-09.09.2020
ATOMS
145564
If the first line in the Lyman series has wavelength $\lambda$, then the first line in Balmer series has the wavelength
1 $\frac{27}{5} \lambda$
2 $\frac{32}{27} \lambda$
3 $\frac{28}{21} \lambda$
4 $\frac{15}{4} \lambda$
Explanation:
A For hydrogen atom, For first line in Lyman series $\mathrm{n}_{1}=1, \text { and } \mathrm{n}_{2}=2$ $\frac{1}{\lambda_{\mathrm{L}}}=\mathrm{R}\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)$ $\frac{1}{\lambda_{\mathrm{L}}}=\mathrm{R}\left(\frac{1}{1}-\frac{1}{4}\right)$ $\frac{1}{\lambda_{\mathrm{L}}}=\frac{3 \mathrm{R}}{4}$ For first line in Balmer series$\mathrm{n}_{1}=2 \text {, and } \mathrm{n}_{2}=3$ $\frac{1}{\lambda_{\mathrm{B}}}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)$ $\frac{1}{\lambda_{\mathrm{B}}}=\mathrm{R}\left(\frac{1}{4}-\frac{1}{9}\right)$ $\frac{1}{\lambda_{\mathrm{B}}}=\mathrm{R}\left(\frac{9-4}{36}\right)$ $\frac{1}{\lambda_{\mathrm{B}}}=\frac{5 \mathrm{R}}{36}$ From equation (i) and equation (ii), we get- $\frac{1}{\lambda_{\mathrm{B}}}=\frac{5 \mathrm{R}}{\frac{1}{36}}$ $\frac{\lambda_{\mathrm{L}}}{4 \mathrm{R}}$ $\frac{1 / \lambda_{\mathrm{B}}}{1 / \lambda_{\mathrm{L}}}=\frac{5 \mathrm{R}}{36} \times \frac{4}{3 \mathrm{R}}$ $\frac{\lambda_{L}}{\lambda_{\mathrm{B}}}=\frac{5 \mathrm{R}}{27 \mathrm{R}}$ $\lambda_{\mathrm{B}}=\frac{27}{5} \lambda_{l}$ $\left(\lambda_{\mathrm{L}}=\lambda\right)$ $\lambda_{\mathrm{B}}=\frac{27}{5} \lambda$
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 $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=\frac{5 \mathrm{R}}{36}$ For second line Balmer series. $\frac{1}{4861}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{4^{2}}\right)=\frac{3 \mathrm{R}}{16}$ divide equation (ii) by (i) then $\frac{\lambda}{4861}=\frac{3 \mathrm{R}}{16} \times \frac{36}{5 \mathrm{R}}$ $\lambda=4861 \times \frac{27}{20}=6563 \AA$
AP EAMCET (18.09.2020) Shift-II
ATOMS
145561
If $\lambda_{1}$ and $\lambda_{2}$ are the wavelengths of the first members of the Lyman and Paschen series respectively, then $\lambda_{1}: \lambda_{2}$ is
1 $1: 3$
2 $1: 30$
3 $7: 50$
4 $7: 108$
Explanation:
D In the Lyman series, the first line is $\mathrm{n}_{1}=1$ and $\mathrm{n}_{2}=2$ $\frac{1}{\lambda_{1}}=\mathrm{R}\left[\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right]$ $\frac{1}{\lambda_{1}}=\mathrm{R}\left[1-\frac{1}{2^{2}}\right]$ $\frac{1}{\lambda_{1}}=\mathrm{R}\left[1-\frac{1}{4}\right]$ $\frac{1}{\lambda_{1}}=\frac{3 \mathrm{R}}{4}$ $\lambda_{1}=\frac{4}{3 \mathrm{R}}$ Similarly in Paschen series the first line is $\mathrm{n}_{1}=3 \text { and } \mathrm{n}_{2}=4$ $\frac{1}{\lambda_{2}}=\mathrm{R}\left[\frac{1}{3^{2}}-\frac{1}{4^{2}}\right]$ $\frac{1}{\lambda_{2}}=\mathrm{R}\left[\frac{1}{9}-\frac{1}{16}\right]$ $\frac{1}{\lambda_{2}}=\frac{7}{144} \mathrm{R}$ $\lambda_{2}=\frac{144}{7 \mathrm{R}}$ From (i) and (ii) $\frac{\lambda_{1}}{\lambda_{2}}=\frac{4}{3 \mathrm{R}} \times \frac{7 \mathrm{R}}{144}$ $\frac{\lambda_{1}}{\lambda_{2}}=\frac{7}{108}$ $\therefore \quad \lambda_{1}: \lambda_{2}=7: 108$
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 $2^{\text {nd }}$ orbital to $3^{\text {rd }}$ orbit is $47.2 \mathrm{eV}$, find the atomic number of the given atom.
1 3
2 4
3 5
4 6
Explanation:
C Given that, $\Delta \mathrm{E}=47.2 \mathrm{eV}, \mathrm{n}_{1}=2, \mathrm{n}_{2}=3$ We know that, $\Delta \mathrm{E}=13.6 \mathrm{Z}^{2}\left(\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right)$ $\Delta \mathrm{E}=13.6 \times \mathrm{Z}^{2} \times\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)$ $47.2=13.6 \mathrm{Z}^{2} \times \frac{5}{36}$ $\mathrm{Z}^{2}=\frac{47.2}{13.6} \times \frac{36}{5}=25$ $\mathrm{Z}=5$
AP EAMCET (18.09.2020) Shift-I
ATOMS
145563
In hydrogen spectrum, if the shortest wavelength in Balmer series is $\lambda$, the shortest wavelength in Brackett series is
1 $\lambda$
2 $\lambda / 2$
3 $4 \lambda$
4 $9 \lambda$
Explanation:
C Shortest wavelength for electron transit from $\infty$ to $\mathrm{n}$ $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{\mathrm{n}^{2}}\right)$ For Balmer series, $\mathrm{n}=2$ $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{2^{2}}\right)$ $\frac{1}{\lambda}=\frac{\mathrm{R}}{4} \Rightarrow \mathrm{R}=\frac{4}{\lambda}$ For Brackett series, $n=4$ $\frac{1}{\lambda(\mathrm{b})}=\mathrm{R}\left(\frac{1}{\mathrm{n}^{2}}\right)=\frac{\mathrm{R}}{16}$ $\frac{1}{\lambda(\mathrm{b})}=\frac{\mathrm{R}}{16}$ \(\frac{16}{\lambda(\mathrm{b})}=\mathrm{R}\) \(\frac{16}{\lambda(\mathrm{b})}=\frac{4}{\lambda}\) \(4 \lambda(\mathrm{b})=16 \lambda\) \(\lambda(\mathrm{b})=4 \lambda\)
TS- EAMCET-09.09.2020
ATOMS
145564
If the first line in the Lyman series has wavelength $\lambda$, then the first line in Balmer series has the wavelength
1 $\frac{27}{5} \lambda$
2 $\frac{32}{27} \lambda$
3 $\frac{28}{21} \lambda$
4 $\frac{15}{4} \lambda$
Explanation:
A For hydrogen atom, For first line in Lyman series $\mathrm{n}_{1}=1, \text { and } \mathrm{n}_{2}=2$ $\frac{1}{\lambda_{\mathrm{L}}}=\mathrm{R}\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)$ $\frac{1}{\lambda_{\mathrm{L}}}=\mathrm{R}\left(\frac{1}{1}-\frac{1}{4}\right)$ $\frac{1}{\lambda_{\mathrm{L}}}=\frac{3 \mathrm{R}}{4}$ For first line in Balmer series$\mathrm{n}_{1}=2 \text {, and } \mathrm{n}_{2}=3$ $\frac{1}{\lambda_{\mathrm{B}}}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)$ $\frac{1}{\lambda_{\mathrm{B}}}=\mathrm{R}\left(\frac{1}{4}-\frac{1}{9}\right)$ $\frac{1}{\lambda_{\mathrm{B}}}=\mathrm{R}\left(\frac{9-4}{36}\right)$ $\frac{1}{\lambda_{\mathrm{B}}}=\frac{5 \mathrm{R}}{36}$ From equation (i) and equation (ii), we get- $\frac{1}{\lambda_{\mathrm{B}}}=\frac{5 \mathrm{R}}{\frac{1}{36}}$ $\frac{\lambda_{\mathrm{L}}}{4 \mathrm{R}}$ $\frac{1 / \lambda_{\mathrm{B}}}{1 / \lambda_{\mathrm{L}}}=\frac{5 \mathrm{R}}{36} \times \frac{4}{3 \mathrm{R}}$ $\frac{\lambda_{L}}{\lambda_{\mathrm{B}}}=\frac{5 \mathrm{R}}{27 \mathrm{R}}$ $\lambda_{\mathrm{B}}=\frac{27}{5} \lambda_{l}$ $\left(\lambda_{\mathrm{L}}=\lambda\right)$ $\lambda_{\mathrm{B}}=\frac{27}{5} \lambda$
