145533
The ratio of speed of electrons in the first excited state of hydrogen atom to the speed of light in vacuum is . Given, Planck's constant $=6.625 \times 10^{-34} \mathrm{~J} . \mathrm{s}$ and permittivity of free space is $8.85 \times 10^{-12} \mathrm{~F}^{-1}$
1 $5 \times 10^{-3}$
2 $7.3 \times 10^{-3}$
3 $3.6 \times 10^{-3}$
4 $36.5 \times 10^{-3}$
Explanation:
D Speed of electron in $\mathrm{n}^{\text {th }}$ orbit for hydrogen atom $\mathrm{v}=\frac{\mathrm{e}^{2}}{2 \varepsilon_{0} \mathrm{nh}}$ For ground state, $\mathrm{n}=1$ and first excited state, $\mathrm{n}=2$ then first excited state velocity of electron $\frac{\mathrm{v}}{\mathrm{c}}=\frac{\left(1.6 \times 10^{-19}\right)^{2}}{2 \times 8.85 \times 10^{-12} \times 3 \times 10^{8} \times 6.6 \times 10^{-34} \times 2}$ $\frac{\mathrm{v}}{\mathrm{c}}=\frac{25.6 \times 10^{-38}}{700.92 \times 10^{-38}}$ $\frac{\mathrm{v}}{\mathrm{c}}=36.5 \times 10^{-3}$
AP EAMCET-25.08.2021
ATOMS
145534
A hydrogen like atom emits radiation of frequency $2.7 \times 10^{15} \mathrm{~Hz}$ when is undergoes a transition from $n=2$ to $n=1$. For the same atom, find frequency of the radiation emitted when a transition occurs from $\mathbf{n}=\mathbf{3}$ to $\mathbf{n}=1$.
1 $3.2 \times 10^{15} \mathrm{~Hz}$
2 $1.8 \times 10^{15} \mathrm{~Hz}$
3 $6.9 \times 10^{15} \mathrm{~Hz}$
4 $4.7 \times 10^{15} \mathrm{~Hz}$
Explanation:
A Given that, frequency $v_{2 \rightarrow 1}=2.7 \times 10^{15} \mathrm{~Hz}$ Then, frequency $v_{3 \rightarrow 1}$ will be We know that, $v_{2 \rightarrow 1}=\frac{13.6 \mathrm{Z}^{2}}{\mathrm{~h}}\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)$ $\frac{13.6 \mathrm{Z}^{2}}{\mathrm{~h}} \times \frac{3}{4}=2.7 \times 10^{15} \mathrm{~Hz}$ When transition from $n=3$ to $n=1$ Then, $\quad v_{3 \rightarrow 1}=\frac{13.6 \mathrm{Z}^{2}}{\mathrm{~h}}\left(1-\frac{1}{9}\right)$ $=\frac{8}{9} \times \frac{13.6 Z^{2}}{h}$ $=\frac{8}{9} \times \frac{4}{3} \times 2.7 \times 10^{15} \mathrm{~Hz}$ $v_{3 \rightarrow 1}=3.2 \times 10^{15} \mathrm{~Hz}$
AP EAMCET-05.10.2021
ATOMS
145537
Hydrogen atom in the ground state absorbs $\Delta E$ amount of energy. If the orbital angular momentum of the electron is increased by $\frac{h}{2 \pi}(h \equiv$ Plank constant $)$, then the magnitude of $\Delta \mathrm{E}$ is
1 $12.09 \mathrm{eV}$
2 $12.75 \mathrm{eV}$
3 $10.2 \mathrm{eV}$
4 $13.6 \mathrm{eV}$
Explanation:
C If angular momentum is increase by $\left(\frac{\mathrm{h}}{2 \pi}\right)=1.05 \times 10^{-34}$ Then angular momentum $\mathrm{L}_{2}-\mathrm{L}_{1}=\frac{2 \mathrm{~h}}{2 \pi}-\frac{\mathrm{h}}{2 \pi}=\frac{\mathrm{h}}{2 \pi}$ So, $\quad \mathrm{n}=2$ Then energy in hydrogen atom $\mathrm{n}=2$ $\mathrm{E}=\frac{-13.5}{\mathrm{n}^{2}}=\frac{-13.6}{(2)^{2}}=3.4 \mathrm{eV}$ Therefore, energy absorbed by electron $=$ Excited energy - ground energy $-3.4 \mathrm{eV}-(-13.6)=\Delta \mathrm{E}$ $\Delta \mathrm{E}=10.2 \mathrm{eV}$
TSEAMCET 04.08.2021
ATOMS
145538
In hydrogen atom spectra, if the ratio of wavelengths corresponding to the first of Lyman series and the first line of Balmer series is $9 \alpha$, the value of $\alpha$ is
1 0.5
2 0.8
3 0.6
4 0.4
Explanation:
C We know that, $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right)$ For first Lyman series $\mathrm{n}=1$ to $\mathrm{n}=2$ $\frac{1}{\lambda_{\mathrm{L}}}=\mathrm{R}\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)$ For first Balmer series $\mathrm{n}=2$ to $\mathrm{n}=3$ $\frac{1}{\lambda_{B}}=R\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)$ $\text { Then } \frac{\lambda_{B}}{\lambda_{L}}=\frac{27}{5}=5$ $9 \alpha=5.4$ $\alpha=0.6$
TSEAMCET 06.08.2021
ATOMS
145539
A mono chromatic radiation of wave length $\lambda$ is incident on a hydrogen sample in ground state the sample subsequently emits radiation of six different wave lengths, then the value of $\lambda$ is [Use ch $=1242 \mathrm{ev}-\mathrm{m}$ ]
1 $80 \mathrm{~nm}$
2 $85.5 \mathrm{~nm}$
3 $97.4 \mathrm{~nm}$
4 $100.2 \mathrm{~nm}$
Explanation:
C Given that, the radiation emit 6 different wavelength, $\text { So, } \quad \frac{\mathrm{n}(\mathrm{n}-1)}{2}=6$ $\mathrm{n}^{2}-\mathrm{n}-12=0$ $\mathrm{n}^{2}-4 \mathrm{n}+3 \mathrm{n}-12=0$ $(\mathrm{n}-4)(\mathrm{n}+3)=0$ $\mathrm{n}=4$ Then wavelength corresponding $\mathrm{n}=4$ $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{1^{2}}-\frac{1}{4^{2}}\right)=\mathrm{R}\left(1-\frac{1}{16}\right)$ $\frac{1}{\lambda}=\frac{15 \mathrm{R}}{16}$ $\lambda=\frac{16}{15 \mathrm{R}}=\frac{16}{15 \times 109677} \mathrm{~cm}$ $\lambda=97.4 \mathrm{~nm}$
145533
The ratio of speed of electrons in the first excited state of hydrogen atom to the speed of light in vacuum is . Given, Planck's constant $=6.625 \times 10^{-34} \mathrm{~J} . \mathrm{s}$ and permittivity of free space is $8.85 \times 10^{-12} \mathrm{~F}^{-1}$
1 $5 \times 10^{-3}$
2 $7.3 \times 10^{-3}$
3 $3.6 \times 10^{-3}$
4 $36.5 \times 10^{-3}$
Explanation:
D Speed of electron in $\mathrm{n}^{\text {th }}$ orbit for hydrogen atom $\mathrm{v}=\frac{\mathrm{e}^{2}}{2 \varepsilon_{0} \mathrm{nh}}$ For ground state, $\mathrm{n}=1$ and first excited state, $\mathrm{n}=2$ then first excited state velocity of electron $\frac{\mathrm{v}}{\mathrm{c}}=\frac{\left(1.6 \times 10^{-19}\right)^{2}}{2 \times 8.85 \times 10^{-12} \times 3 \times 10^{8} \times 6.6 \times 10^{-34} \times 2}$ $\frac{\mathrm{v}}{\mathrm{c}}=\frac{25.6 \times 10^{-38}}{700.92 \times 10^{-38}}$ $\frac{\mathrm{v}}{\mathrm{c}}=36.5 \times 10^{-3}$
AP EAMCET-25.08.2021
ATOMS
145534
A hydrogen like atom emits radiation of frequency $2.7 \times 10^{15} \mathrm{~Hz}$ when is undergoes a transition from $n=2$ to $n=1$. For the same atom, find frequency of the radiation emitted when a transition occurs from $\mathbf{n}=\mathbf{3}$ to $\mathbf{n}=1$.
1 $3.2 \times 10^{15} \mathrm{~Hz}$
2 $1.8 \times 10^{15} \mathrm{~Hz}$
3 $6.9 \times 10^{15} \mathrm{~Hz}$
4 $4.7 \times 10^{15} \mathrm{~Hz}$
Explanation:
A Given that, frequency $v_{2 \rightarrow 1}=2.7 \times 10^{15} \mathrm{~Hz}$ Then, frequency $v_{3 \rightarrow 1}$ will be We know that, $v_{2 \rightarrow 1}=\frac{13.6 \mathrm{Z}^{2}}{\mathrm{~h}}\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)$ $\frac{13.6 \mathrm{Z}^{2}}{\mathrm{~h}} \times \frac{3}{4}=2.7 \times 10^{15} \mathrm{~Hz}$ When transition from $n=3$ to $n=1$ Then, $\quad v_{3 \rightarrow 1}=\frac{13.6 \mathrm{Z}^{2}}{\mathrm{~h}}\left(1-\frac{1}{9}\right)$ $=\frac{8}{9} \times \frac{13.6 Z^{2}}{h}$ $=\frac{8}{9} \times \frac{4}{3} \times 2.7 \times 10^{15} \mathrm{~Hz}$ $v_{3 \rightarrow 1}=3.2 \times 10^{15} \mathrm{~Hz}$
AP EAMCET-05.10.2021
ATOMS
145537
Hydrogen atom in the ground state absorbs $\Delta E$ amount of energy. If the orbital angular momentum of the electron is increased by $\frac{h}{2 \pi}(h \equiv$ Plank constant $)$, then the magnitude of $\Delta \mathrm{E}$ is
1 $12.09 \mathrm{eV}$
2 $12.75 \mathrm{eV}$
3 $10.2 \mathrm{eV}$
4 $13.6 \mathrm{eV}$
Explanation:
C If angular momentum is increase by $\left(\frac{\mathrm{h}}{2 \pi}\right)=1.05 \times 10^{-34}$ Then angular momentum $\mathrm{L}_{2}-\mathrm{L}_{1}=\frac{2 \mathrm{~h}}{2 \pi}-\frac{\mathrm{h}}{2 \pi}=\frac{\mathrm{h}}{2 \pi}$ So, $\quad \mathrm{n}=2$ Then energy in hydrogen atom $\mathrm{n}=2$ $\mathrm{E}=\frac{-13.5}{\mathrm{n}^{2}}=\frac{-13.6}{(2)^{2}}=3.4 \mathrm{eV}$ Therefore, energy absorbed by electron $=$ Excited energy - ground energy $-3.4 \mathrm{eV}-(-13.6)=\Delta \mathrm{E}$ $\Delta \mathrm{E}=10.2 \mathrm{eV}$
TSEAMCET 04.08.2021
ATOMS
145538
In hydrogen atom spectra, if the ratio of wavelengths corresponding to the first of Lyman series and the first line of Balmer series is $9 \alpha$, the value of $\alpha$ is
1 0.5
2 0.8
3 0.6
4 0.4
Explanation:
C We know that, $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right)$ For first Lyman series $\mathrm{n}=1$ to $\mathrm{n}=2$ $\frac{1}{\lambda_{\mathrm{L}}}=\mathrm{R}\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)$ For first Balmer series $\mathrm{n}=2$ to $\mathrm{n}=3$ $\frac{1}{\lambda_{B}}=R\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)$ $\text { Then } \frac{\lambda_{B}}{\lambda_{L}}=\frac{27}{5}=5$ $9 \alpha=5.4$ $\alpha=0.6$
TSEAMCET 06.08.2021
ATOMS
145539
A mono chromatic radiation of wave length $\lambda$ is incident on a hydrogen sample in ground state the sample subsequently emits radiation of six different wave lengths, then the value of $\lambda$ is [Use ch $=1242 \mathrm{ev}-\mathrm{m}$ ]
1 $80 \mathrm{~nm}$
2 $85.5 \mathrm{~nm}$
3 $97.4 \mathrm{~nm}$
4 $100.2 \mathrm{~nm}$
Explanation:
C Given that, the radiation emit 6 different wavelength, $\text { So, } \quad \frac{\mathrm{n}(\mathrm{n}-1)}{2}=6$ $\mathrm{n}^{2}-\mathrm{n}-12=0$ $\mathrm{n}^{2}-4 \mathrm{n}+3 \mathrm{n}-12=0$ $(\mathrm{n}-4)(\mathrm{n}+3)=0$ $\mathrm{n}=4$ Then wavelength corresponding $\mathrm{n}=4$ $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{1^{2}}-\frac{1}{4^{2}}\right)=\mathrm{R}\left(1-\frac{1}{16}\right)$ $\frac{1}{\lambda}=\frac{15 \mathrm{R}}{16}$ $\lambda=\frac{16}{15 \mathrm{R}}=\frac{16}{15 \times 109677} \mathrm{~cm}$ $\lambda=97.4 \mathrm{~nm}$
145533
The ratio of speed of electrons in the first excited state of hydrogen atom to the speed of light in vacuum is . Given, Planck's constant $=6.625 \times 10^{-34} \mathrm{~J} . \mathrm{s}$ and permittivity of free space is $8.85 \times 10^{-12} \mathrm{~F}^{-1}$
1 $5 \times 10^{-3}$
2 $7.3 \times 10^{-3}$
3 $3.6 \times 10^{-3}$
4 $36.5 \times 10^{-3}$
Explanation:
D Speed of electron in $\mathrm{n}^{\text {th }}$ orbit for hydrogen atom $\mathrm{v}=\frac{\mathrm{e}^{2}}{2 \varepsilon_{0} \mathrm{nh}}$ For ground state, $\mathrm{n}=1$ and first excited state, $\mathrm{n}=2$ then first excited state velocity of electron $\frac{\mathrm{v}}{\mathrm{c}}=\frac{\left(1.6 \times 10^{-19}\right)^{2}}{2 \times 8.85 \times 10^{-12} \times 3 \times 10^{8} \times 6.6 \times 10^{-34} \times 2}$ $\frac{\mathrm{v}}{\mathrm{c}}=\frac{25.6 \times 10^{-38}}{700.92 \times 10^{-38}}$ $\frac{\mathrm{v}}{\mathrm{c}}=36.5 \times 10^{-3}$
AP EAMCET-25.08.2021
ATOMS
145534
A hydrogen like atom emits radiation of frequency $2.7 \times 10^{15} \mathrm{~Hz}$ when is undergoes a transition from $n=2$ to $n=1$. For the same atom, find frequency of the radiation emitted when a transition occurs from $\mathbf{n}=\mathbf{3}$ to $\mathbf{n}=1$.
1 $3.2 \times 10^{15} \mathrm{~Hz}$
2 $1.8 \times 10^{15} \mathrm{~Hz}$
3 $6.9 \times 10^{15} \mathrm{~Hz}$
4 $4.7 \times 10^{15} \mathrm{~Hz}$
Explanation:
A Given that, frequency $v_{2 \rightarrow 1}=2.7 \times 10^{15} \mathrm{~Hz}$ Then, frequency $v_{3 \rightarrow 1}$ will be We know that, $v_{2 \rightarrow 1}=\frac{13.6 \mathrm{Z}^{2}}{\mathrm{~h}}\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)$ $\frac{13.6 \mathrm{Z}^{2}}{\mathrm{~h}} \times \frac{3}{4}=2.7 \times 10^{15} \mathrm{~Hz}$ When transition from $n=3$ to $n=1$ Then, $\quad v_{3 \rightarrow 1}=\frac{13.6 \mathrm{Z}^{2}}{\mathrm{~h}}\left(1-\frac{1}{9}\right)$ $=\frac{8}{9} \times \frac{13.6 Z^{2}}{h}$ $=\frac{8}{9} \times \frac{4}{3} \times 2.7 \times 10^{15} \mathrm{~Hz}$ $v_{3 \rightarrow 1}=3.2 \times 10^{15} \mathrm{~Hz}$
AP EAMCET-05.10.2021
ATOMS
145537
Hydrogen atom in the ground state absorbs $\Delta E$ amount of energy. If the orbital angular momentum of the electron is increased by $\frac{h}{2 \pi}(h \equiv$ Plank constant $)$, then the magnitude of $\Delta \mathrm{E}$ is
1 $12.09 \mathrm{eV}$
2 $12.75 \mathrm{eV}$
3 $10.2 \mathrm{eV}$
4 $13.6 \mathrm{eV}$
Explanation:
C If angular momentum is increase by $\left(\frac{\mathrm{h}}{2 \pi}\right)=1.05 \times 10^{-34}$ Then angular momentum $\mathrm{L}_{2}-\mathrm{L}_{1}=\frac{2 \mathrm{~h}}{2 \pi}-\frac{\mathrm{h}}{2 \pi}=\frac{\mathrm{h}}{2 \pi}$ So, $\quad \mathrm{n}=2$ Then energy in hydrogen atom $\mathrm{n}=2$ $\mathrm{E}=\frac{-13.5}{\mathrm{n}^{2}}=\frac{-13.6}{(2)^{2}}=3.4 \mathrm{eV}$ Therefore, energy absorbed by electron $=$ Excited energy - ground energy $-3.4 \mathrm{eV}-(-13.6)=\Delta \mathrm{E}$ $\Delta \mathrm{E}=10.2 \mathrm{eV}$
TSEAMCET 04.08.2021
ATOMS
145538
In hydrogen atom spectra, if the ratio of wavelengths corresponding to the first of Lyman series and the first line of Balmer series is $9 \alpha$, the value of $\alpha$ is
1 0.5
2 0.8
3 0.6
4 0.4
Explanation:
C We know that, $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right)$ For first Lyman series $\mathrm{n}=1$ to $\mathrm{n}=2$ $\frac{1}{\lambda_{\mathrm{L}}}=\mathrm{R}\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)$ For first Balmer series $\mathrm{n}=2$ to $\mathrm{n}=3$ $\frac{1}{\lambda_{B}}=R\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)$ $\text { Then } \frac{\lambda_{B}}{\lambda_{L}}=\frac{27}{5}=5$ $9 \alpha=5.4$ $\alpha=0.6$
TSEAMCET 06.08.2021
ATOMS
145539
A mono chromatic radiation of wave length $\lambda$ is incident on a hydrogen sample in ground state the sample subsequently emits radiation of six different wave lengths, then the value of $\lambda$ is [Use ch $=1242 \mathrm{ev}-\mathrm{m}$ ]
1 $80 \mathrm{~nm}$
2 $85.5 \mathrm{~nm}$
3 $97.4 \mathrm{~nm}$
4 $100.2 \mathrm{~nm}$
Explanation:
C Given that, the radiation emit 6 different wavelength, $\text { So, } \quad \frac{\mathrm{n}(\mathrm{n}-1)}{2}=6$ $\mathrm{n}^{2}-\mathrm{n}-12=0$ $\mathrm{n}^{2}-4 \mathrm{n}+3 \mathrm{n}-12=0$ $(\mathrm{n}-4)(\mathrm{n}+3)=0$ $\mathrm{n}=4$ Then wavelength corresponding $\mathrm{n}=4$ $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{1^{2}}-\frac{1}{4^{2}}\right)=\mathrm{R}\left(1-\frac{1}{16}\right)$ $\frac{1}{\lambda}=\frac{15 \mathrm{R}}{16}$ $\lambda=\frac{16}{15 \mathrm{R}}=\frac{16}{15 \times 109677} \mathrm{~cm}$ $\lambda=97.4 \mathrm{~nm}$
145533
The ratio of speed of electrons in the first excited state of hydrogen atom to the speed of light in vacuum is . Given, Planck's constant $=6.625 \times 10^{-34} \mathrm{~J} . \mathrm{s}$ and permittivity of free space is $8.85 \times 10^{-12} \mathrm{~F}^{-1}$
1 $5 \times 10^{-3}$
2 $7.3 \times 10^{-3}$
3 $3.6 \times 10^{-3}$
4 $36.5 \times 10^{-3}$
Explanation:
D Speed of electron in $\mathrm{n}^{\text {th }}$ orbit for hydrogen atom $\mathrm{v}=\frac{\mathrm{e}^{2}}{2 \varepsilon_{0} \mathrm{nh}}$ For ground state, $\mathrm{n}=1$ and first excited state, $\mathrm{n}=2$ then first excited state velocity of electron $\frac{\mathrm{v}}{\mathrm{c}}=\frac{\left(1.6 \times 10^{-19}\right)^{2}}{2 \times 8.85 \times 10^{-12} \times 3 \times 10^{8} \times 6.6 \times 10^{-34} \times 2}$ $\frac{\mathrm{v}}{\mathrm{c}}=\frac{25.6 \times 10^{-38}}{700.92 \times 10^{-38}}$ $\frac{\mathrm{v}}{\mathrm{c}}=36.5 \times 10^{-3}$
AP EAMCET-25.08.2021
ATOMS
145534
A hydrogen like atom emits radiation of frequency $2.7 \times 10^{15} \mathrm{~Hz}$ when is undergoes a transition from $n=2$ to $n=1$. For the same atom, find frequency of the radiation emitted when a transition occurs from $\mathbf{n}=\mathbf{3}$ to $\mathbf{n}=1$.
1 $3.2 \times 10^{15} \mathrm{~Hz}$
2 $1.8 \times 10^{15} \mathrm{~Hz}$
3 $6.9 \times 10^{15} \mathrm{~Hz}$
4 $4.7 \times 10^{15} \mathrm{~Hz}$
Explanation:
A Given that, frequency $v_{2 \rightarrow 1}=2.7 \times 10^{15} \mathrm{~Hz}$ Then, frequency $v_{3 \rightarrow 1}$ will be We know that, $v_{2 \rightarrow 1}=\frac{13.6 \mathrm{Z}^{2}}{\mathrm{~h}}\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)$ $\frac{13.6 \mathrm{Z}^{2}}{\mathrm{~h}} \times \frac{3}{4}=2.7 \times 10^{15} \mathrm{~Hz}$ When transition from $n=3$ to $n=1$ Then, $\quad v_{3 \rightarrow 1}=\frac{13.6 \mathrm{Z}^{2}}{\mathrm{~h}}\left(1-\frac{1}{9}\right)$ $=\frac{8}{9} \times \frac{13.6 Z^{2}}{h}$ $=\frac{8}{9} \times \frac{4}{3} \times 2.7 \times 10^{15} \mathrm{~Hz}$ $v_{3 \rightarrow 1}=3.2 \times 10^{15} \mathrm{~Hz}$
AP EAMCET-05.10.2021
ATOMS
145537
Hydrogen atom in the ground state absorbs $\Delta E$ amount of energy. If the orbital angular momentum of the electron is increased by $\frac{h}{2 \pi}(h \equiv$ Plank constant $)$, then the magnitude of $\Delta \mathrm{E}$ is
1 $12.09 \mathrm{eV}$
2 $12.75 \mathrm{eV}$
3 $10.2 \mathrm{eV}$
4 $13.6 \mathrm{eV}$
Explanation:
C If angular momentum is increase by $\left(\frac{\mathrm{h}}{2 \pi}\right)=1.05 \times 10^{-34}$ Then angular momentum $\mathrm{L}_{2}-\mathrm{L}_{1}=\frac{2 \mathrm{~h}}{2 \pi}-\frac{\mathrm{h}}{2 \pi}=\frac{\mathrm{h}}{2 \pi}$ So, $\quad \mathrm{n}=2$ Then energy in hydrogen atom $\mathrm{n}=2$ $\mathrm{E}=\frac{-13.5}{\mathrm{n}^{2}}=\frac{-13.6}{(2)^{2}}=3.4 \mathrm{eV}$ Therefore, energy absorbed by electron $=$ Excited energy - ground energy $-3.4 \mathrm{eV}-(-13.6)=\Delta \mathrm{E}$ $\Delta \mathrm{E}=10.2 \mathrm{eV}$
TSEAMCET 04.08.2021
ATOMS
145538
In hydrogen atom spectra, if the ratio of wavelengths corresponding to the first of Lyman series and the first line of Balmer series is $9 \alpha$, the value of $\alpha$ is
1 0.5
2 0.8
3 0.6
4 0.4
Explanation:
C We know that, $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right)$ For first Lyman series $\mathrm{n}=1$ to $\mathrm{n}=2$ $\frac{1}{\lambda_{\mathrm{L}}}=\mathrm{R}\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)$ For first Balmer series $\mathrm{n}=2$ to $\mathrm{n}=3$ $\frac{1}{\lambda_{B}}=R\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)$ $\text { Then } \frac{\lambda_{B}}{\lambda_{L}}=\frac{27}{5}=5$ $9 \alpha=5.4$ $\alpha=0.6$
TSEAMCET 06.08.2021
ATOMS
145539
A mono chromatic radiation of wave length $\lambda$ is incident on a hydrogen sample in ground state the sample subsequently emits radiation of six different wave lengths, then the value of $\lambda$ is [Use ch $=1242 \mathrm{ev}-\mathrm{m}$ ]
1 $80 \mathrm{~nm}$
2 $85.5 \mathrm{~nm}$
3 $97.4 \mathrm{~nm}$
4 $100.2 \mathrm{~nm}$
Explanation:
C Given that, the radiation emit 6 different wavelength, $\text { So, } \quad \frac{\mathrm{n}(\mathrm{n}-1)}{2}=6$ $\mathrm{n}^{2}-\mathrm{n}-12=0$ $\mathrm{n}^{2}-4 \mathrm{n}+3 \mathrm{n}-12=0$ $(\mathrm{n}-4)(\mathrm{n}+3)=0$ $\mathrm{n}=4$ Then wavelength corresponding $\mathrm{n}=4$ $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{1^{2}}-\frac{1}{4^{2}}\right)=\mathrm{R}\left(1-\frac{1}{16}\right)$ $\frac{1}{\lambda}=\frac{15 \mathrm{R}}{16}$ $\lambda=\frac{16}{15 \mathrm{R}}=\frac{16}{15 \times 109677} \mathrm{~cm}$ $\lambda=97.4 \mathrm{~nm}$
145533
The ratio of speed of electrons in the first excited state of hydrogen atom to the speed of light in vacuum is . Given, Planck's constant $=6.625 \times 10^{-34} \mathrm{~J} . \mathrm{s}$ and permittivity of free space is $8.85 \times 10^{-12} \mathrm{~F}^{-1}$
1 $5 \times 10^{-3}$
2 $7.3 \times 10^{-3}$
3 $3.6 \times 10^{-3}$
4 $36.5 \times 10^{-3}$
Explanation:
D Speed of electron in $\mathrm{n}^{\text {th }}$ orbit for hydrogen atom $\mathrm{v}=\frac{\mathrm{e}^{2}}{2 \varepsilon_{0} \mathrm{nh}}$ For ground state, $\mathrm{n}=1$ and first excited state, $\mathrm{n}=2$ then first excited state velocity of electron $\frac{\mathrm{v}}{\mathrm{c}}=\frac{\left(1.6 \times 10^{-19}\right)^{2}}{2 \times 8.85 \times 10^{-12} \times 3 \times 10^{8} \times 6.6 \times 10^{-34} \times 2}$ $\frac{\mathrm{v}}{\mathrm{c}}=\frac{25.6 \times 10^{-38}}{700.92 \times 10^{-38}}$ $\frac{\mathrm{v}}{\mathrm{c}}=36.5 \times 10^{-3}$
AP EAMCET-25.08.2021
ATOMS
145534
A hydrogen like atom emits radiation of frequency $2.7 \times 10^{15} \mathrm{~Hz}$ when is undergoes a transition from $n=2$ to $n=1$. For the same atom, find frequency of the radiation emitted when a transition occurs from $\mathbf{n}=\mathbf{3}$ to $\mathbf{n}=1$.
1 $3.2 \times 10^{15} \mathrm{~Hz}$
2 $1.8 \times 10^{15} \mathrm{~Hz}$
3 $6.9 \times 10^{15} \mathrm{~Hz}$
4 $4.7 \times 10^{15} \mathrm{~Hz}$
Explanation:
A Given that, frequency $v_{2 \rightarrow 1}=2.7 \times 10^{15} \mathrm{~Hz}$ Then, frequency $v_{3 \rightarrow 1}$ will be We know that, $v_{2 \rightarrow 1}=\frac{13.6 \mathrm{Z}^{2}}{\mathrm{~h}}\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)$ $\frac{13.6 \mathrm{Z}^{2}}{\mathrm{~h}} \times \frac{3}{4}=2.7 \times 10^{15} \mathrm{~Hz}$ When transition from $n=3$ to $n=1$ Then, $\quad v_{3 \rightarrow 1}=\frac{13.6 \mathrm{Z}^{2}}{\mathrm{~h}}\left(1-\frac{1}{9}\right)$ $=\frac{8}{9} \times \frac{13.6 Z^{2}}{h}$ $=\frac{8}{9} \times \frac{4}{3} \times 2.7 \times 10^{15} \mathrm{~Hz}$ $v_{3 \rightarrow 1}=3.2 \times 10^{15} \mathrm{~Hz}$
AP EAMCET-05.10.2021
ATOMS
145537
Hydrogen atom in the ground state absorbs $\Delta E$ amount of energy. If the orbital angular momentum of the electron is increased by $\frac{h}{2 \pi}(h \equiv$ Plank constant $)$, then the magnitude of $\Delta \mathrm{E}$ is
1 $12.09 \mathrm{eV}$
2 $12.75 \mathrm{eV}$
3 $10.2 \mathrm{eV}$
4 $13.6 \mathrm{eV}$
Explanation:
C If angular momentum is increase by $\left(\frac{\mathrm{h}}{2 \pi}\right)=1.05 \times 10^{-34}$ Then angular momentum $\mathrm{L}_{2}-\mathrm{L}_{1}=\frac{2 \mathrm{~h}}{2 \pi}-\frac{\mathrm{h}}{2 \pi}=\frac{\mathrm{h}}{2 \pi}$ So, $\quad \mathrm{n}=2$ Then energy in hydrogen atom $\mathrm{n}=2$ $\mathrm{E}=\frac{-13.5}{\mathrm{n}^{2}}=\frac{-13.6}{(2)^{2}}=3.4 \mathrm{eV}$ Therefore, energy absorbed by electron $=$ Excited energy - ground energy $-3.4 \mathrm{eV}-(-13.6)=\Delta \mathrm{E}$ $\Delta \mathrm{E}=10.2 \mathrm{eV}$
TSEAMCET 04.08.2021
ATOMS
145538
In hydrogen atom spectra, if the ratio of wavelengths corresponding to the first of Lyman series and the first line of Balmer series is $9 \alpha$, the value of $\alpha$ is
1 0.5
2 0.8
3 0.6
4 0.4
Explanation:
C We know that, $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right)$ For first Lyman series $\mathrm{n}=1$ to $\mathrm{n}=2$ $\frac{1}{\lambda_{\mathrm{L}}}=\mathrm{R}\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)$ For first Balmer series $\mathrm{n}=2$ to $\mathrm{n}=3$ $\frac{1}{\lambda_{B}}=R\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)$ $\text { Then } \frac{\lambda_{B}}{\lambda_{L}}=\frac{27}{5}=5$ $9 \alpha=5.4$ $\alpha=0.6$
TSEAMCET 06.08.2021
ATOMS
145539
A mono chromatic radiation of wave length $\lambda$ is incident on a hydrogen sample in ground state the sample subsequently emits radiation of six different wave lengths, then the value of $\lambda$ is [Use ch $=1242 \mathrm{ev}-\mathrm{m}$ ]
1 $80 \mathrm{~nm}$
2 $85.5 \mathrm{~nm}$
3 $97.4 \mathrm{~nm}$
4 $100.2 \mathrm{~nm}$
Explanation:
C Given that, the radiation emit 6 different wavelength, $\text { So, } \quad \frac{\mathrm{n}(\mathrm{n}-1)}{2}=6$ $\mathrm{n}^{2}-\mathrm{n}-12=0$ $\mathrm{n}^{2}-4 \mathrm{n}+3 \mathrm{n}-12=0$ $(\mathrm{n}-4)(\mathrm{n}+3)=0$ $\mathrm{n}=4$ Then wavelength corresponding $\mathrm{n}=4$ $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{1^{2}}-\frac{1}{4^{2}}\right)=\mathrm{R}\left(1-\frac{1}{16}\right)$ $\frac{1}{\lambda}=\frac{15 \mathrm{R}}{16}$ $\lambda=\frac{16}{15 \mathrm{R}}=\frac{16}{15 \times 109677} \mathrm{~cm}$ $\lambda=97.4 \mathrm{~nm}$
