NEET Test Series from KOTA - 10 Papers In MS WORD
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ATOMS
145540
What is the shortest wavelength present in the Balmer series of spectral line? [Where $R$ is Rydberg constant]
1 $\frac{1}{\mathrm{R}}$
2 $\frac{3}{\mathrm{R}}$
3 $\frac{2}{\mathrm{R}}$
4 $\frac{4}{\mathrm{R}}$
Explanation:
D Wavelength for Balmer series is $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{\mathrm{n}^{2}}\right)$ $n=\infty$, the limit of the series observed. $\therefore \quad\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{4}-\frac{1}{\infty}\right)$ $\frac{1}{\lambda}=\frac{\mathrm{R}}{4}$ $\therefore \quad\lambda=\frac{4}{\mathrm{R}}$
GUJCET-PCE- 2021
ATOMS
145541
Hydrogen atoms in its ground state are excited by monochromatic radiation of photon energy $12.8 \mathrm{eV}$. If the ionization potential of hydrogen atom is $13.6 \mathrm{eV}$, the number of spectral lines emitted according to Bohr theory will be
1 6
2 4
3 3
4 1
Explanation:
C Ionization energy corresponding to ionization potential $=-13.6 \mathrm{eV}$ photon energy incident $=12.1 \mathrm{eV}$ So the energy of electron in excited state $=-13.6+$ 12.1 $=-1.5 \mathrm{eV}$ $\mathrm{E}_{\mathrm{n}}=\frac{-13.6}{\mathrm{n}^{2}} \mathrm{eV}$ $-1.5=\frac{-13.6}{\mathrm{n}^{2}}$ $\mathrm{n}^{2}=\frac{-13.6}{-1.5} \approx 9$ $\mathrm{n}=3$
UPSEE 2020
ATOMS
145542
An electron makes a transition from outer orbit $(n=4)$ to the inner orbit $(n=2)$ of a hydrogen atom. The wave number of the emitted radiation is:
1 $\frac{2 \mathrm{R}}{16}$
2 $\frac{3 R}{16}$
3 $\frac{4 \mathrm{R}}{16}$
4 $\frac{5 \mathrm{R}}{16}$
Explanation:
B When transition of hydrogen atom from orbit $\mathrm{n}=4$ to $\mathrm{n}=2$ Then, $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right)$ $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{4^{2}}\right)$ $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{4}-\frac{1}{16}\right)=\frac{3 \mathrm{R}}{16}$ $\frac{1}{\lambda}=\frac{3 \mathrm{R}}{16}$ But we know that wave number $v=\frac{1}{\lambda}$ So, $\quad v=\frac{1}{\lambda}=\frac{3 \mathrm{R}}{16}$
AP EAMCET-23.09.2020
ATOMS
145543
The wavelength of the first line of Balmer series of hydrogen atom is $\lambda$, what will be the wavelength of the same line in doubly ionized lithium?
1 $\frac{\lambda}{2}$
2 $\frac{\lambda}{9}$
3 $\frac{\lambda}{8}$
4 $\frac{\lambda}{27}$
Explanation:
B Given that, for first Balmer line or hydrogen atom $\frac{1}{\lambda}=\mathrm{RZ}^{2}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)$ For hydrogen $\mathrm{Z}=1$ then, $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)$ The wavelength for doubly ionization lithium atom in same line $\mathrm{Z}=3$ $\frac{1}{\lambda_{\mathrm{L}}}=\mathrm{RZ}^{2}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)$ $\frac{1}{\lambda_{\mathrm{L}}}=\mathrm{R}(3)^{2}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)$ Then, $\frac{1}{\lambda_{\mathrm{L}}}=9 \mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)$ Then ratio, $\frac{\frac{1}{\lambda}}{\frac{1}{\lambda_{\mathrm{L}}}}=\frac{\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)}{9 \mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)}$ $\frac{\lambda_{\mathrm{L}}}{\lambda}=\frac{1}{9}$ $\lambda_{\mathrm{L}}=\frac{\lambda}{9}$
145540
What is the shortest wavelength present in the Balmer series of spectral line? [Where $R$ is Rydberg constant]
1 $\frac{1}{\mathrm{R}}$
2 $\frac{3}{\mathrm{R}}$
3 $\frac{2}{\mathrm{R}}$
4 $\frac{4}{\mathrm{R}}$
Explanation:
D Wavelength for Balmer series is $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{\mathrm{n}^{2}}\right)$ $n=\infty$, the limit of the series observed. $\therefore \quad\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{4}-\frac{1}{\infty}\right)$ $\frac{1}{\lambda}=\frac{\mathrm{R}}{4}$ $\therefore \quad\lambda=\frac{4}{\mathrm{R}}$
GUJCET-PCE- 2021
ATOMS
145541
Hydrogen atoms in its ground state are excited by monochromatic radiation of photon energy $12.8 \mathrm{eV}$. If the ionization potential of hydrogen atom is $13.6 \mathrm{eV}$, the number of spectral lines emitted according to Bohr theory will be
1 6
2 4
3 3
4 1
Explanation:
C Ionization energy corresponding to ionization potential $=-13.6 \mathrm{eV}$ photon energy incident $=12.1 \mathrm{eV}$ So the energy of electron in excited state $=-13.6+$ 12.1 $=-1.5 \mathrm{eV}$ $\mathrm{E}_{\mathrm{n}}=\frac{-13.6}{\mathrm{n}^{2}} \mathrm{eV}$ $-1.5=\frac{-13.6}{\mathrm{n}^{2}}$ $\mathrm{n}^{2}=\frac{-13.6}{-1.5} \approx 9$ $\mathrm{n}=3$
UPSEE 2020
ATOMS
145542
An electron makes a transition from outer orbit $(n=4)$ to the inner orbit $(n=2)$ of a hydrogen atom. The wave number of the emitted radiation is:
1 $\frac{2 \mathrm{R}}{16}$
2 $\frac{3 R}{16}$
3 $\frac{4 \mathrm{R}}{16}$
4 $\frac{5 \mathrm{R}}{16}$
Explanation:
B When transition of hydrogen atom from orbit $\mathrm{n}=4$ to $\mathrm{n}=2$ Then, $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right)$ $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{4^{2}}\right)$ $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{4}-\frac{1}{16}\right)=\frac{3 \mathrm{R}}{16}$ $\frac{1}{\lambda}=\frac{3 \mathrm{R}}{16}$ But we know that wave number $v=\frac{1}{\lambda}$ So, $\quad v=\frac{1}{\lambda}=\frac{3 \mathrm{R}}{16}$
AP EAMCET-23.09.2020
ATOMS
145543
The wavelength of the first line of Balmer series of hydrogen atom is $\lambda$, what will be the wavelength of the same line in doubly ionized lithium?
1 $\frac{\lambda}{2}$
2 $\frac{\lambda}{9}$
3 $\frac{\lambda}{8}$
4 $\frac{\lambda}{27}$
Explanation:
B Given that, for first Balmer line or hydrogen atom $\frac{1}{\lambda}=\mathrm{RZ}^{2}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)$ For hydrogen $\mathrm{Z}=1$ then, $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)$ The wavelength for doubly ionization lithium atom in same line $\mathrm{Z}=3$ $\frac{1}{\lambda_{\mathrm{L}}}=\mathrm{RZ}^{2}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)$ $\frac{1}{\lambda_{\mathrm{L}}}=\mathrm{R}(3)^{2}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)$ Then, $\frac{1}{\lambda_{\mathrm{L}}}=9 \mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)$ Then ratio, $\frac{\frac{1}{\lambda}}{\frac{1}{\lambda_{\mathrm{L}}}}=\frac{\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)}{9 \mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)}$ $\frac{\lambda_{\mathrm{L}}}{\lambda}=\frac{1}{9}$ $\lambda_{\mathrm{L}}=\frac{\lambda}{9}$
145540
What is the shortest wavelength present in the Balmer series of spectral line? [Where $R$ is Rydberg constant]
1 $\frac{1}{\mathrm{R}}$
2 $\frac{3}{\mathrm{R}}$
3 $\frac{2}{\mathrm{R}}$
4 $\frac{4}{\mathrm{R}}$
Explanation:
D Wavelength for Balmer series is $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{\mathrm{n}^{2}}\right)$ $n=\infty$, the limit of the series observed. $\therefore \quad\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{4}-\frac{1}{\infty}\right)$ $\frac{1}{\lambda}=\frac{\mathrm{R}}{4}$ $\therefore \quad\lambda=\frac{4}{\mathrm{R}}$
GUJCET-PCE- 2021
ATOMS
145541
Hydrogen atoms in its ground state are excited by monochromatic radiation of photon energy $12.8 \mathrm{eV}$. If the ionization potential of hydrogen atom is $13.6 \mathrm{eV}$, the number of spectral lines emitted according to Bohr theory will be
1 6
2 4
3 3
4 1
Explanation:
C Ionization energy corresponding to ionization potential $=-13.6 \mathrm{eV}$ photon energy incident $=12.1 \mathrm{eV}$ So the energy of electron in excited state $=-13.6+$ 12.1 $=-1.5 \mathrm{eV}$ $\mathrm{E}_{\mathrm{n}}=\frac{-13.6}{\mathrm{n}^{2}} \mathrm{eV}$ $-1.5=\frac{-13.6}{\mathrm{n}^{2}}$ $\mathrm{n}^{2}=\frac{-13.6}{-1.5} \approx 9$ $\mathrm{n}=3$
UPSEE 2020
ATOMS
145542
An electron makes a transition from outer orbit $(n=4)$ to the inner orbit $(n=2)$ of a hydrogen atom. The wave number of the emitted radiation is:
1 $\frac{2 \mathrm{R}}{16}$
2 $\frac{3 R}{16}$
3 $\frac{4 \mathrm{R}}{16}$
4 $\frac{5 \mathrm{R}}{16}$
Explanation:
B When transition of hydrogen atom from orbit $\mathrm{n}=4$ to $\mathrm{n}=2$ Then, $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right)$ $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{4^{2}}\right)$ $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{4}-\frac{1}{16}\right)=\frac{3 \mathrm{R}}{16}$ $\frac{1}{\lambda}=\frac{3 \mathrm{R}}{16}$ But we know that wave number $v=\frac{1}{\lambda}$ So, $\quad v=\frac{1}{\lambda}=\frac{3 \mathrm{R}}{16}$
AP EAMCET-23.09.2020
ATOMS
145543
The wavelength of the first line of Balmer series of hydrogen atom is $\lambda$, what will be the wavelength of the same line in doubly ionized lithium?
1 $\frac{\lambda}{2}$
2 $\frac{\lambda}{9}$
3 $\frac{\lambda}{8}$
4 $\frac{\lambda}{27}$
Explanation:
B Given that, for first Balmer line or hydrogen atom $\frac{1}{\lambda}=\mathrm{RZ}^{2}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)$ For hydrogen $\mathrm{Z}=1$ then, $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)$ The wavelength for doubly ionization lithium atom in same line $\mathrm{Z}=3$ $\frac{1}{\lambda_{\mathrm{L}}}=\mathrm{RZ}^{2}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)$ $\frac{1}{\lambda_{\mathrm{L}}}=\mathrm{R}(3)^{2}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)$ Then, $\frac{1}{\lambda_{\mathrm{L}}}=9 \mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)$ Then ratio, $\frac{\frac{1}{\lambda}}{\frac{1}{\lambda_{\mathrm{L}}}}=\frac{\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)}{9 \mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)}$ $\frac{\lambda_{\mathrm{L}}}{\lambda}=\frac{1}{9}$ $\lambda_{\mathrm{L}}=\frac{\lambda}{9}$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
ATOMS
145540
What is the shortest wavelength present in the Balmer series of spectral line? [Where $R$ is Rydberg constant]
1 $\frac{1}{\mathrm{R}}$
2 $\frac{3}{\mathrm{R}}$
3 $\frac{2}{\mathrm{R}}$
4 $\frac{4}{\mathrm{R}}$
Explanation:
D Wavelength for Balmer series is $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{\mathrm{n}^{2}}\right)$ $n=\infty$, the limit of the series observed. $\therefore \quad\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{4}-\frac{1}{\infty}\right)$ $\frac{1}{\lambda}=\frac{\mathrm{R}}{4}$ $\therefore \quad\lambda=\frac{4}{\mathrm{R}}$
GUJCET-PCE- 2021
ATOMS
145541
Hydrogen atoms in its ground state are excited by monochromatic radiation of photon energy $12.8 \mathrm{eV}$. If the ionization potential of hydrogen atom is $13.6 \mathrm{eV}$, the number of spectral lines emitted according to Bohr theory will be
1 6
2 4
3 3
4 1
Explanation:
C Ionization energy corresponding to ionization potential $=-13.6 \mathrm{eV}$ photon energy incident $=12.1 \mathrm{eV}$ So the energy of electron in excited state $=-13.6+$ 12.1 $=-1.5 \mathrm{eV}$ $\mathrm{E}_{\mathrm{n}}=\frac{-13.6}{\mathrm{n}^{2}} \mathrm{eV}$ $-1.5=\frac{-13.6}{\mathrm{n}^{2}}$ $\mathrm{n}^{2}=\frac{-13.6}{-1.5} \approx 9$ $\mathrm{n}=3$
UPSEE 2020
ATOMS
145542
An electron makes a transition from outer orbit $(n=4)$ to the inner orbit $(n=2)$ of a hydrogen atom. The wave number of the emitted radiation is:
1 $\frac{2 \mathrm{R}}{16}$
2 $\frac{3 R}{16}$
3 $\frac{4 \mathrm{R}}{16}$
4 $\frac{5 \mathrm{R}}{16}$
Explanation:
B When transition of hydrogen atom from orbit $\mathrm{n}=4$ to $\mathrm{n}=2$ Then, $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right)$ $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{4^{2}}\right)$ $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{4}-\frac{1}{16}\right)=\frac{3 \mathrm{R}}{16}$ $\frac{1}{\lambda}=\frac{3 \mathrm{R}}{16}$ But we know that wave number $v=\frac{1}{\lambda}$ So, $\quad v=\frac{1}{\lambda}=\frac{3 \mathrm{R}}{16}$
AP EAMCET-23.09.2020
ATOMS
145543
The wavelength of the first line of Balmer series of hydrogen atom is $\lambda$, what will be the wavelength of the same line in doubly ionized lithium?
1 $\frac{\lambda}{2}$
2 $\frac{\lambda}{9}$
3 $\frac{\lambda}{8}$
4 $\frac{\lambda}{27}$
Explanation:
B Given that, for first Balmer line or hydrogen atom $\frac{1}{\lambda}=\mathrm{RZ}^{2}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)$ For hydrogen $\mathrm{Z}=1$ then, $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)$ The wavelength for doubly ionization lithium atom in same line $\mathrm{Z}=3$ $\frac{1}{\lambda_{\mathrm{L}}}=\mathrm{RZ}^{2}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)$ $\frac{1}{\lambda_{\mathrm{L}}}=\mathrm{R}(3)^{2}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)$ Then, $\frac{1}{\lambda_{\mathrm{L}}}=9 \mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)$ Then ratio, $\frac{\frac{1}{\lambda}}{\frac{1}{\lambda_{\mathrm{L}}}}=\frac{\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)}{9 \mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)}$ $\frac{\lambda_{\mathrm{L}}}{\lambda}=\frac{1}{9}$ $\lambda_{\mathrm{L}}=\frac{\lambda}{9}$