NEET Test Series from KOTA - 10 Papers In MS WORD
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ATOMS
145745
Electron in Hydrogen atom first jumps from third excited state to second excited state and then from second excited state to first excited state. The ratio of the wavelengths $\lambda_{1}: \lambda_{2}$ emitted in the two cases respectively is
1 $\frac{27}{5}$
2 $\frac{7}{5}$
3 $\frac{27}{20}$
4 $\frac{20}{7}$
Explanation:
A We know that, $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{\mathrm{n}_{2}^{2}}-\frac{1}{\mathrm{n}_{1}^{2}}\right)$ For $\mathrm{n}_{1}=2$ and $\mathrm{n}_{2}=3$ $\frac{1}{\lambda_{1}}=\mathrm{R}\left(\frac{1}{3^{2}}-\frac{1}{2^{2}}\right)$ $\frac{1}{\lambda_{1}}=\mathrm{R}\left(\frac{1}{9}-\frac{1}{4}\right)$ Or $\quad \lambda_{1}=\frac{36}{5 \mathrm{R}}$ For $\mathrm{n}_{1}=2$ and $\mathrm{n}_{2}=1$ $\frac{1}{\lambda_{2}}=\mathrm{R}\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)$ $\frac{1}{\lambda_{2}}=\mathrm{R}\left(\frac{4-1}{4}\right)$ $\lambda_{2}=\frac{4}{3 \mathrm{R}}$ Dividing equation (i) by equation (ii), we get- $\therefore \quad \frac{\lambda_{1}}{\lambda_{2}}=\frac{36}{5 \mathrm{R}} \times \frac{3 \mathrm{R}}{4}$ $\frac{\lambda_{1}}{\lambda_{2}}=\frac{27}{5}$
MHT-CET 2020
ATOMS
145747
If the wavelength of a photon emitted due to transition of election from third orbit to first orbit in a hydrogen atom is $\lambda$, then the wavelength of a photon emitted due to transition of electro from fourth orbit to second orbit will be
1 $\frac{25}{9} \lambda$
2 $\frac{128}{27} \lambda$
3 $\frac{36}{7} \lambda$
4 $\frac{125}{11} \lambda$
Explanation:
B We know that, Emitted wavelength, $\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right]$ Transition from third to first, $\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{(1)^{2}}-\frac{1}{(3)^{2}}\right]$ $\frac{1}{\lambda}=\mathrm{R}\left[1-\frac{1}{9}\right]$ $\frac{1}{\lambda}=\mathrm{R}\left[\frac{9-1}{9}\right]$ $\frac{1}{\lambda}=\frac{8 \mathrm{R}}{9}$ $\lambda=\frac{9}{8 \mathrm{R}}$ Now, transition from fourth to second, $\frac{1}{\lambda^{\prime}}=\mathrm{R}\left[\frac{1}{(2)^{2}}-\frac{1}{(4)^{2}}\right]$ $\frac{1}{\lambda^{\prime}}=\mathrm{R}\left[\frac{1}{4}-\frac{1}{16}\right]$ $\frac{1}{\lambda^{\prime}}=\mathrm{R}\left[\frac{4-1}{16}\right]=\frac{3 \mathrm{R}}{16}$ $\lambda^{\prime}=\frac{16}{3 \mathrm{R}}$ Wavelength of photon due to transition of electron from fourth orbit to second orbit is. Equation (ii) divided by equation (i), we get- $\frac{\lambda^{\prime}}{\lambda}=\frac{\frac{16}{3 \mathrm{R}}}{\frac{9}{8 \mathrm{R}}}=\frac{16 \times 8}{3 \times 9}=\frac{128}{27}$ $\lambda^{\prime}=\frac{128}{27} \lambda$
AP EAMCET-24.04.2019
ATOMS
145748
When a hydrogen atom emits a photon during a transition from $n=4$ to $n=2$, its recoil speed is about,
1 $4.28 \mathrm{~ms}^{-1}$
2 $0.814 \mathrm{~ms}^{-1}$
3 $2.07 \mathrm{~ms}^{-1}$
4 $0.407 \mathrm{~ms}^{-1}$
Explanation:
B We know that, momentum of photon, $\mathrm{p}=\frac{\mathrm{h}}{\lambda}$ $\mathrm{mv}=\frac{\mathrm{h}}{\lambda} \quad(\therefore \mathrm{p}=\mathrm{mv})$ $\mathrm{v}=\frac{\mathrm{h}}{\mathrm{m} \lambda}$ Hydrogen atom emits a photon during transition from- $\therefore \quad \mathrm{n}_{2}=4 \text { to } \mathrm{n}_{1}=2$ $\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right]$ $\frac{1}{\lambda}=1.097 \times 10^{7}\left[\frac{1}{(2)^{2}}-\frac{1}{(4)^{2}}\right]$ $\lambda=4.86 \times 10^{-7} \mathrm{~m}$ $\therefore$ Mass of photon $(\mathrm{m})=1.6 \times 10^{-27} \mathrm{~kg}$ $\mathrm{v}=\frac{6.62 \times 10^{-34}}{1.6 \times 10^{-27} \times 4.86 \times 10^{-7}}$ $\mathrm{v}=0.8513 \mathrm{~m} / \mathrm{sec}$ $\mathrm{v}\approx 0.814 \mathrm{~m} / \mathrm{sec}$
AP EAMCET (21.04.2019) Shift-II
ATOMS
145749
A hydrogen atom emits a photon of wavelength $\frac{36}{35 R}$ when it is jumped form its $n$th excited state. Then the quantum number $n$ is ( $R$ is Rydberg constant)
1 8
2 7
3 5
4 6
Explanation:
D When a electron jump from $\mathrm{n}^{\text {th }}$ to ground state, then wavelength is given by. $\lambda_{\mathrm{n}^{\text {th }}}=\frac{36}{35 \mathrm{R}}$ Wavelength in Lyman series $\frac{1}{\lambda_{\mathrm{L}_{y}}}=\mathrm{R}\left[1-\frac{1}{\mathrm{n}^{2}}\right]$ $\frac{1}{\lambda_{\mathrm{L}_{\mathrm{y}}}}=\mathrm{R}\left[\frac{\mathrm{n}^{2}-1}{\mathrm{n}^{2}}\right]$ $\lambda_{\mathrm{L}_{\mathrm{y}}}=\frac{\mathrm{n}^{2}}{\mathrm{R}\left[\mathrm{n}^{2}-1\right]}$ Hence, $\lambda_{\mathrm{L}_{\mathrm{y}}}=\lambda_{\mathrm{n}^{\mathrm{th}}}$ From equation (i) and equation (ii), we get- $\frac{\mathrm{n}^{2}}{\mathrm{R}\left[\mathrm{n}^{2}-1\right]}=\frac{36}{35 \mathrm{R}}$ $\frac{\mathrm{n}^{2}}{\mathrm{n}^{2}-1}=\frac{36}{35}$ $35 \mathrm{n}^{2}=36 \mathrm{n}^{2}-36$ $36 \mathrm{n}^{2}-35 \mathrm{n}^{2}=36$ $\mathrm{n}^{2}=36$ $\mathrm{n}=6 \text { state }$
145745
Electron in Hydrogen atom first jumps from third excited state to second excited state and then from second excited state to first excited state. The ratio of the wavelengths $\lambda_{1}: \lambda_{2}$ emitted in the two cases respectively is
1 $\frac{27}{5}$
2 $\frac{7}{5}$
3 $\frac{27}{20}$
4 $\frac{20}{7}$
Explanation:
A We know that, $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{\mathrm{n}_{2}^{2}}-\frac{1}{\mathrm{n}_{1}^{2}}\right)$ For $\mathrm{n}_{1}=2$ and $\mathrm{n}_{2}=3$ $\frac{1}{\lambda_{1}}=\mathrm{R}\left(\frac{1}{3^{2}}-\frac{1}{2^{2}}\right)$ $\frac{1}{\lambda_{1}}=\mathrm{R}\left(\frac{1}{9}-\frac{1}{4}\right)$ Or $\quad \lambda_{1}=\frac{36}{5 \mathrm{R}}$ For $\mathrm{n}_{1}=2$ and $\mathrm{n}_{2}=1$ $\frac{1}{\lambda_{2}}=\mathrm{R}\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)$ $\frac{1}{\lambda_{2}}=\mathrm{R}\left(\frac{4-1}{4}\right)$ $\lambda_{2}=\frac{4}{3 \mathrm{R}}$ Dividing equation (i) by equation (ii), we get- $\therefore \quad \frac{\lambda_{1}}{\lambda_{2}}=\frac{36}{5 \mathrm{R}} \times \frac{3 \mathrm{R}}{4}$ $\frac{\lambda_{1}}{\lambda_{2}}=\frac{27}{5}$
MHT-CET 2020
ATOMS
145747
If the wavelength of a photon emitted due to transition of election from third orbit to first orbit in a hydrogen atom is $\lambda$, then the wavelength of a photon emitted due to transition of electro from fourth orbit to second orbit will be
1 $\frac{25}{9} \lambda$
2 $\frac{128}{27} \lambda$
3 $\frac{36}{7} \lambda$
4 $\frac{125}{11} \lambda$
Explanation:
B We know that, Emitted wavelength, $\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right]$ Transition from third to first, $\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{(1)^{2}}-\frac{1}{(3)^{2}}\right]$ $\frac{1}{\lambda}=\mathrm{R}\left[1-\frac{1}{9}\right]$ $\frac{1}{\lambda}=\mathrm{R}\left[\frac{9-1}{9}\right]$ $\frac{1}{\lambda}=\frac{8 \mathrm{R}}{9}$ $\lambda=\frac{9}{8 \mathrm{R}}$ Now, transition from fourth to second, $\frac{1}{\lambda^{\prime}}=\mathrm{R}\left[\frac{1}{(2)^{2}}-\frac{1}{(4)^{2}}\right]$ $\frac{1}{\lambda^{\prime}}=\mathrm{R}\left[\frac{1}{4}-\frac{1}{16}\right]$ $\frac{1}{\lambda^{\prime}}=\mathrm{R}\left[\frac{4-1}{16}\right]=\frac{3 \mathrm{R}}{16}$ $\lambda^{\prime}=\frac{16}{3 \mathrm{R}}$ Wavelength of photon due to transition of electron from fourth orbit to second orbit is. Equation (ii) divided by equation (i), we get- $\frac{\lambda^{\prime}}{\lambda}=\frac{\frac{16}{3 \mathrm{R}}}{\frac{9}{8 \mathrm{R}}}=\frac{16 \times 8}{3 \times 9}=\frac{128}{27}$ $\lambda^{\prime}=\frac{128}{27} \lambda$
AP EAMCET-24.04.2019
ATOMS
145748
When a hydrogen atom emits a photon during a transition from $n=4$ to $n=2$, its recoil speed is about,
1 $4.28 \mathrm{~ms}^{-1}$
2 $0.814 \mathrm{~ms}^{-1}$
3 $2.07 \mathrm{~ms}^{-1}$
4 $0.407 \mathrm{~ms}^{-1}$
Explanation:
B We know that, momentum of photon, $\mathrm{p}=\frac{\mathrm{h}}{\lambda}$ $\mathrm{mv}=\frac{\mathrm{h}}{\lambda} \quad(\therefore \mathrm{p}=\mathrm{mv})$ $\mathrm{v}=\frac{\mathrm{h}}{\mathrm{m} \lambda}$ Hydrogen atom emits a photon during transition from- $\therefore \quad \mathrm{n}_{2}=4 \text { to } \mathrm{n}_{1}=2$ $\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right]$ $\frac{1}{\lambda}=1.097 \times 10^{7}\left[\frac{1}{(2)^{2}}-\frac{1}{(4)^{2}}\right]$ $\lambda=4.86 \times 10^{-7} \mathrm{~m}$ $\therefore$ Mass of photon $(\mathrm{m})=1.6 \times 10^{-27} \mathrm{~kg}$ $\mathrm{v}=\frac{6.62 \times 10^{-34}}{1.6 \times 10^{-27} \times 4.86 \times 10^{-7}}$ $\mathrm{v}=0.8513 \mathrm{~m} / \mathrm{sec}$ $\mathrm{v}\approx 0.814 \mathrm{~m} / \mathrm{sec}$
AP EAMCET (21.04.2019) Shift-II
ATOMS
145749
A hydrogen atom emits a photon of wavelength $\frac{36}{35 R}$ when it is jumped form its $n$th excited state. Then the quantum number $n$ is ( $R$ is Rydberg constant)
1 8
2 7
3 5
4 6
Explanation:
D When a electron jump from $\mathrm{n}^{\text {th }}$ to ground state, then wavelength is given by. $\lambda_{\mathrm{n}^{\text {th }}}=\frac{36}{35 \mathrm{R}}$ Wavelength in Lyman series $\frac{1}{\lambda_{\mathrm{L}_{y}}}=\mathrm{R}\left[1-\frac{1}{\mathrm{n}^{2}}\right]$ $\frac{1}{\lambda_{\mathrm{L}_{\mathrm{y}}}}=\mathrm{R}\left[\frac{\mathrm{n}^{2}-1}{\mathrm{n}^{2}}\right]$ $\lambda_{\mathrm{L}_{\mathrm{y}}}=\frac{\mathrm{n}^{2}}{\mathrm{R}\left[\mathrm{n}^{2}-1\right]}$ Hence, $\lambda_{\mathrm{L}_{\mathrm{y}}}=\lambda_{\mathrm{n}^{\mathrm{th}}}$ From equation (i) and equation (ii), we get- $\frac{\mathrm{n}^{2}}{\mathrm{R}\left[\mathrm{n}^{2}-1\right]}=\frac{36}{35 \mathrm{R}}$ $\frac{\mathrm{n}^{2}}{\mathrm{n}^{2}-1}=\frac{36}{35}$ $35 \mathrm{n}^{2}=36 \mathrm{n}^{2}-36$ $36 \mathrm{n}^{2}-35 \mathrm{n}^{2}=36$ $\mathrm{n}^{2}=36$ $\mathrm{n}=6 \text { state }$
145745
Electron in Hydrogen atom first jumps from third excited state to second excited state and then from second excited state to first excited state. The ratio of the wavelengths $\lambda_{1}: \lambda_{2}$ emitted in the two cases respectively is
1 $\frac{27}{5}$
2 $\frac{7}{5}$
3 $\frac{27}{20}$
4 $\frac{20}{7}$
Explanation:
A We know that, $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{\mathrm{n}_{2}^{2}}-\frac{1}{\mathrm{n}_{1}^{2}}\right)$ For $\mathrm{n}_{1}=2$ and $\mathrm{n}_{2}=3$ $\frac{1}{\lambda_{1}}=\mathrm{R}\left(\frac{1}{3^{2}}-\frac{1}{2^{2}}\right)$ $\frac{1}{\lambda_{1}}=\mathrm{R}\left(\frac{1}{9}-\frac{1}{4}\right)$ Or $\quad \lambda_{1}=\frac{36}{5 \mathrm{R}}$ For $\mathrm{n}_{1}=2$ and $\mathrm{n}_{2}=1$ $\frac{1}{\lambda_{2}}=\mathrm{R}\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)$ $\frac{1}{\lambda_{2}}=\mathrm{R}\left(\frac{4-1}{4}\right)$ $\lambda_{2}=\frac{4}{3 \mathrm{R}}$ Dividing equation (i) by equation (ii), we get- $\therefore \quad \frac{\lambda_{1}}{\lambda_{2}}=\frac{36}{5 \mathrm{R}} \times \frac{3 \mathrm{R}}{4}$ $\frac{\lambda_{1}}{\lambda_{2}}=\frac{27}{5}$
MHT-CET 2020
ATOMS
145747
If the wavelength of a photon emitted due to transition of election from third orbit to first orbit in a hydrogen atom is $\lambda$, then the wavelength of a photon emitted due to transition of electro from fourth orbit to second orbit will be
1 $\frac{25}{9} \lambda$
2 $\frac{128}{27} \lambda$
3 $\frac{36}{7} \lambda$
4 $\frac{125}{11} \lambda$
Explanation:
B We know that, Emitted wavelength, $\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right]$ Transition from third to first, $\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{(1)^{2}}-\frac{1}{(3)^{2}}\right]$ $\frac{1}{\lambda}=\mathrm{R}\left[1-\frac{1}{9}\right]$ $\frac{1}{\lambda}=\mathrm{R}\left[\frac{9-1}{9}\right]$ $\frac{1}{\lambda}=\frac{8 \mathrm{R}}{9}$ $\lambda=\frac{9}{8 \mathrm{R}}$ Now, transition from fourth to second, $\frac{1}{\lambda^{\prime}}=\mathrm{R}\left[\frac{1}{(2)^{2}}-\frac{1}{(4)^{2}}\right]$ $\frac{1}{\lambda^{\prime}}=\mathrm{R}\left[\frac{1}{4}-\frac{1}{16}\right]$ $\frac{1}{\lambda^{\prime}}=\mathrm{R}\left[\frac{4-1}{16}\right]=\frac{3 \mathrm{R}}{16}$ $\lambda^{\prime}=\frac{16}{3 \mathrm{R}}$ Wavelength of photon due to transition of electron from fourth orbit to second orbit is. Equation (ii) divided by equation (i), we get- $\frac{\lambda^{\prime}}{\lambda}=\frac{\frac{16}{3 \mathrm{R}}}{\frac{9}{8 \mathrm{R}}}=\frac{16 \times 8}{3 \times 9}=\frac{128}{27}$ $\lambda^{\prime}=\frac{128}{27} \lambda$
AP EAMCET-24.04.2019
ATOMS
145748
When a hydrogen atom emits a photon during a transition from $n=4$ to $n=2$, its recoil speed is about,
1 $4.28 \mathrm{~ms}^{-1}$
2 $0.814 \mathrm{~ms}^{-1}$
3 $2.07 \mathrm{~ms}^{-1}$
4 $0.407 \mathrm{~ms}^{-1}$
Explanation:
B We know that, momentum of photon, $\mathrm{p}=\frac{\mathrm{h}}{\lambda}$ $\mathrm{mv}=\frac{\mathrm{h}}{\lambda} \quad(\therefore \mathrm{p}=\mathrm{mv})$ $\mathrm{v}=\frac{\mathrm{h}}{\mathrm{m} \lambda}$ Hydrogen atom emits a photon during transition from- $\therefore \quad \mathrm{n}_{2}=4 \text { to } \mathrm{n}_{1}=2$ $\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right]$ $\frac{1}{\lambda}=1.097 \times 10^{7}\left[\frac{1}{(2)^{2}}-\frac{1}{(4)^{2}}\right]$ $\lambda=4.86 \times 10^{-7} \mathrm{~m}$ $\therefore$ Mass of photon $(\mathrm{m})=1.6 \times 10^{-27} \mathrm{~kg}$ $\mathrm{v}=\frac{6.62 \times 10^{-34}}{1.6 \times 10^{-27} \times 4.86 \times 10^{-7}}$ $\mathrm{v}=0.8513 \mathrm{~m} / \mathrm{sec}$ $\mathrm{v}\approx 0.814 \mathrm{~m} / \mathrm{sec}$
AP EAMCET (21.04.2019) Shift-II
ATOMS
145749
A hydrogen atom emits a photon of wavelength $\frac{36}{35 R}$ when it is jumped form its $n$th excited state. Then the quantum number $n$ is ( $R$ is Rydberg constant)
1 8
2 7
3 5
4 6
Explanation:
D When a electron jump from $\mathrm{n}^{\text {th }}$ to ground state, then wavelength is given by. $\lambda_{\mathrm{n}^{\text {th }}}=\frac{36}{35 \mathrm{R}}$ Wavelength in Lyman series $\frac{1}{\lambda_{\mathrm{L}_{y}}}=\mathrm{R}\left[1-\frac{1}{\mathrm{n}^{2}}\right]$ $\frac{1}{\lambda_{\mathrm{L}_{\mathrm{y}}}}=\mathrm{R}\left[\frac{\mathrm{n}^{2}-1}{\mathrm{n}^{2}}\right]$ $\lambda_{\mathrm{L}_{\mathrm{y}}}=\frac{\mathrm{n}^{2}}{\mathrm{R}\left[\mathrm{n}^{2}-1\right]}$ Hence, $\lambda_{\mathrm{L}_{\mathrm{y}}}=\lambda_{\mathrm{n}^{\mathrm{th}}}$ From equation (i) and equation (ii), we get- $\frac{\mathrm{n}^{2}}{\mathrm{R}\left[\mathrm{n}^{2}-1\right]}=\frac{36}{35 \mathrm{R}}$ $\frac{\mathrm{n}^{2}}{\mathrm{n}^{2}-1}=\frac{36}{35}$ $35 \mathrm{n}^{2}=36 \mathrm{n}^{2}-36$ $36 \mathrm{n}^{2}-35 \mathrm{n}^{2}=36$ $\mathrm{n}^{2}=36$ $\mathrm{n}=6 \text { state }$
145745
Electron in Hydrogen atom first jumps from third excited state to second excited state and then from second excited state to first excited state. The ratio of the wavelengths $\lambda_{1}: \lambda_{2}$ emitted in the two cases respectively is
1 $\frac{27}{5}$
2 $\frac{7}{5}$
3 $\frac{27}{20}$
4 $\frac{20}{7}$
Explanation:
A We know that, $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{\mathrm{n}_{2}^{2}}-\frac{1}{\mathrm{n}_{1}^{2}}\right)$ For $\mathrm{n}_{1}=2$ and $\mathrm{n}_{2}=3$ $\frac{1}{\lambda_{1}}=\mathrm{R}\left(\frac{1}{3^{2}}-\frac{1}{2^{2}}\right)$ $\frac{1}{\lambda_{1}}=\mathrm{R}\left(\frac{1}{9}-\frac{1}{4}\right)$ Or $\quad \lambda_{1}=\frac{36}{5 \mathrm{R}}$ For $\mathrm{n}_{1}=2$ and $\mathrm{n}_{2}=1$ $\frac{1}{\lambda_{2}}=\mathrm{R}\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)$ $\frac{1}{\lambda_{2}}=\mathrm{R}\left(\frac{4-1}{4}\right)$ $\lambda_{2}=\frac{4}{3 \mathrm{R}}$ Dividing equation (i) by equation (ii), we get- $\therefore \quad \frac{\lambda_{1}}{\lambda_{2}}=\frac{36}{5 \mathrm{R}} \times \frac{3 \mathrm{R}}{4}$ $\frac{\lambda_{1}}{\lambda_{2}}=\frac{27}{5}$
MHT-CET 2020
ATOMS
145747
If the wavelength of a photon emitted due to transition of election from third orbit to first orbit in a hydrogen atom is $\lambda$, then the wavelength of a photon emitted due to transition of electro from fourth orbit to second orbit will be
1 $\frac{25}{9} \lambda$
2 $\frac{128}{27} \lambda$
3 $\frac{36}{7} \lambda$
4 $\frac{125}{11} \lambda$
Explanation:
B We know that, Emitted wavelength, $\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right]$ Transition from third to first, $\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{(1)^{2}}-\frac{1}{(3)^{2}}\right]$ $\frac{1}{\lambda}=\mathrm{R}\left[1-\frac{1}{9}\right]$ $\frac{1}{\lambda}=\mathrm{R}\left[\frac{9-1}{9}\right]$ $\frac{1}{\lambda}=\frac{8 \mathrm{R}}{9}$ $\lambda=\frac{9}{8 \mathrm{R}}$ Now, transition from fourth to second, $\frac{1}{\lambda^{\prime}}=\mathrm{R}\left[\frac{1}{(2)^{2}}-\frac{1}{(4)^{2}}\right]$ $\frac{1}{\lambda^{\prime}}=\mathrm{R}\left[\frac{1}{4}-\frac{1}{16}\right]$ $\frac{1}{\lambda^{\prime}}=\mathrm{R}\left[\frac{4-1}{16}\right]=\frac{3 \mathrm{R}}{16}$ $\lambda^{\prime}=\frac{16}{3 \mathrm{R}}$ Wavelength of photon due to transition of electron from fourth orbit to second orbit is. Equation (ii) divided by equation (i), we get- $\frac{\lambda^{\prime}}{\lambda}=\frac{\frac{16}{3 \mathrm{R}}}{\frac{9}{8 \mathrm{R}}}=\frac{16 \times 8}{3 \times 9}=\frac{128}{27}$ $\lambda^{\prime}=\frac{128}{27} \lambda$
AP EAMCET-24.04.2019
ATOMS
145748
When a hydrogen atom emits a photon during a transition from $n=4$ to $n=2$, its recoil speed is about,
1 $4.28 \mathrm{~ms}^{-1}$
2 $0.814 \mathrm{~ms}^{-1}$
3 $2.07 \mathrm{~ms}^{-1}$
4 $0.407 \mathrm{~ms}^{-1}$
Explanation:
B We know that, momentum of photon, $\mathrm{p}=\frac{\mathrm{h}}{\lambda}$ $\mathrm{mv}=\frac{\mathrm{h}}{\lambda} \quad(\therefore \mathrm{p}=\mathrm{mv})$ $\mathrm{v}=\frac{\mathrm{h}}{\mathrm{m} \lambda}$ Hydrogen atom emits a photon during transition from- $\therefore \quad \mathrm{n}_{2}=4 \text { to } \mathrm{n}_{1}=2$ $\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right]$ $\frac{1}{\lambda}=1.097 \times 10^{7}\left[\frac{1}{(2)^{2}}-\frac{1}{(4)^{2}}\right]$ $\lambda=4.86 \times 10^{-7} \mathrm{~m}$ $\therefore$ Mass of photon $(\mathrm{m})=1.6 \times 10^{-27} \mathrm{~kg}$ $\mathrm{v}=\frac{6.62 \times 10^{-34}}{1.6 \times 10^{-27} \times 4.86 \times 10^{-7}}$ $\mathrm{v}=0.8513 \mathrm{~m} / \mathrm{sec}$ $\mathrm{v}\approx 0.814 \mathrm{~m} / \mathrm{sec}$
AP EAMCET (21.04.2019) Shift-II
ATOMS
145749
A hydrogen atom emits a photon of wavelength $\frac{36}{35 R}$ when it is jumped form its $n$th excited state. Then the quantum number $n$ is ( $R$ is Rydberg constant)
1 8
2 7
3 5
4 6
Explanation:
D When a electron jump from $\mathrm{n}^{\text {th }}$ to ground state, then wavelength is given by. $\lambda_{\mathrm{n}^{\text {th }}}=\frac{36}{35 \mathrm{R}}$ Wavelength in Lyman series $\frac{1}{\lambda_{\mathrm{L}_{y}}}=\mathrm{R}\left[1-\frac{1}{\mathrm{n}^{2}}\right]$ $\frac{1}{\lambda_{\mathrm{L}_{\mathrm{y}}}}=\mathrm{R}\left[\frac{\mathrm{n}^{2}-1}{\mathrm{n}^{2}}\right]$ $\lambda_{\mathrm{L}_{\mathrm{y}}}=\frac{\mathrm{n}^{2}}{\mathrm{R}\left[\mathrm{n}^{2}-1\right]}$ Hence, $\lambda_{\mathrm{L}_{\mathrm{y}}}=\lambda_{\mathrm{n}^{\mathrm{th}}}$ From equation (i) and equation (ii), we get- $\frac{\mathrm{n}^{2}}{\mathrm{R}\left[\mathrm{n}^{2}-1\right]}=\frac{36}{35 \mathrm{R}}$ $\frac{\mathrm{n}^{2}}{\mathrm{n}^{2}-1}=\frac{36}{35}$ $35 \mathrm{n}^{2}=36 \mathrm{n}^{2}-36$ $36 \mathrm{n}^{2}-35 \mathrm{n}^{2}=36$ $\mathrm{n}^{2}=36$ $\mathrm{n}=6 \text { state }$
