145735
The energy levels of an hydrogen atom are shown below. The transition corresponding to emission of shortest wavelength is
1 $\mathrm{C}$
2 $\mathrm{D}$
3 $\mathrm{B}$
4 $\mathrm{A}$
Explanation:
B Since we know that, $\Delta \mathrm{E}=\frac{\mathrm{hc}}{\lambda}$ $\lambda \propto \frac{1}{\Delta \mathrm{E}}$ For shortest wavelength, energy gap should be maximum. So, correct choice is transition from $n=3$ to $n=1$.
JEE Main-06.04.2023
ATOMS
145736
Hydrogen atom from excited state comes to the ground by emitting a photon of wavelength $\lambda$. The value of principal quantum number ' $n$ ' of the excited state will be: ( $R$ : Rydberg constant)
B According to Rydberg's formula, $\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{\mathrm{n}_{\mathrm{f}}^{2}}-\frac{1}{\mathrm{n}_{\mathrm{i}}^{2}}\right]$ $\mathrm{n}_{\mathrm{f}}=1$ $\mathrm{n}_{\mathrm{i}}=\mathrm{n}$ $\because \quad \frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{1^{2}}-\frac{1}{\mathrm{n}^{2}}\right]$ $\frac{1}{\lambda}=\mathrm{R}\left[1-\frac{1}{\mathrm{n}^{2}}\right]$ Multiplying by $\lambda$ on both side in equation (i), $1=\lambda \mathrm{R}\left[1-\frac{1}{\mathrm{n}^{2}}\right]$ $\frac{1}{\lambda \mathrm{R}}=1-\frac{1}{\mathrm{n}^{2}}$ $\frac{1}{\mathrm{n}^{2}}=1-\frac{1}{\lambda \mathrm{R}}$ $\frac{1}{\mathrm{n}^{2}}=\frac{\lambda \mathrm{R}-1}{\lambda \mathrm{R}}$ $\mathrm{n}=\sqrt{\frac{\lambda \mathrm{R}}{\lambda \mathrm{R}-1}}$
JEE Main-25.07.2022
ATOMS
145737
The total number of spectral lines observed when electron returns from the $6^{\text {th }}$ shell until the $2^{\text {nd }}$ shell in hydrogen atom is
1 15
2 10
3 20
4 25
Explanation:
B Given, $\mathrm{n}_{1}=2$ and $\mathrm{n}_{2}=6$ Total no. of spectral lines $=\frac{\left(\mathrm{n}_{2}-\mathrm{n}_{1}\right) \cdot\left(\mathrm{n}_{2}-\mathrm{n}_{1}+1\right)}{2}$ $=\frac{(6-2) \cdot(6-2+1)}{2}$ $=\frac{4 \times 5}{2}$ $=10$
TSEAMCET 18.07.2022
ATOMS
145738
The light emitted in the transition $n=3$ to $n=2$ (where $n$ is the principal quantum number of the state) in hydrogen is called $H_{\alpha}$-light find the maximum work function that a metal can have so that $H_{\alpha}$ light can emit photoelectrons from it.
1 $1.5 \mathrm{eV}$
2 $2.89 \mathrm{eV}$
3 $1.89 \mathrm{eV}$
4 $3.5 \mathrm{eV}$
Explanation:
C We have given that $\mathrm{n}_{1}=2$ $\mathrm{n}_{2}=3$ Energy possessed by $\mathrm{H}_{\alpha}$ light $=13.6\left(\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right)$ $=13.6\left[\frac{1}{(2)^{2}}-\frac{1}{(3)^{2}}\right]$ $=13.6\left[\frac{1}{4}-\frac{1}{9}\right]$ $=13.6 \times \frac{5}{36}$ $=\frac{68.0}{36}$ $=1.89 \mathrm{eV}$
TSEAMCET 31.07.2022
ATOMS
145739
An electron in a hydrogen atom makes a transition form $n=n_{1}$ to $n=n_{2}$ (where $n$ is a principal quantum number of a state). The time period of electron in the initial state is eight times than that of the final state, then which of the following statements is TRUE?
1 $\mathrm{n}_{1}=3 \mathrm{n}_{2}$
2 $\mathrm{n}_{1}=4 \mathrm{n}_{2}$
3 $\mathrm{n}_{1}=2 \mathrm{n}_{2}$
4 $\mathrm{n}_{1}=5 \mathrm{n}_{2}$
Explanation:
C The time period of electron in nth orbit $T_{n} \propto n^{3}$ Then, $\mathrm{T}_{1}=\mathrm{Kn}_{1}{ }^{3}$ $\mathrm{~T}_{2}=\mathrm{Kn}_{2}{ }^{3}$ Equation (ii) divided by equation (iii) we have $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\frac{\mathrm{Kn}_{1}^{3}}{\mathrm{Kn}_{2}^{3}}$ $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\left(\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}\right)^{3}$ According to question, $\mathrm{T}_{1}=8 \mathrm{~T}_{2}$ From equation (iv) $\frac{8 \mathrm{~T}_{2}}{\mathrm{~T}_{2}}=\left(\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}\right)^{3}$ $\left(\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}\right)^{3}=8$ $\left(\frac{n_{1}}{n_{2}}\right)^{3}=(2)^{3}$ $\frac{n_{1}}{n_{2}}=2$ $\mathrm{n}_{1}=2 \mathrm{n}_{2}$
145735
The energy levels of an hydrogen atom are shown below. The transition corresponding to emission of shortest wavelength is
1 $\mathrm{C}$
2 $\mathrm{D}$
3 $\mathrm{B}$
4 $\mathrm{A}$
Explanation:
B Since we know that, $\Delta \mathrm{E}=\frac{\mathrm{hc}}{\lambda}$ $\lambda \propto \frac{1}{\Delta \mathrm{E}}$ For shortest wavelength, energy gap should be maximum. So, correct choice is transition from $n=3$ to $n=1$.
JEE Main-06.04.2023
ATOMS
145736
Hydrogen atom from excited state comes to the ground by emitting a photon of wavelength $\lambda$. The value of principal quantum number ' $n$ ' of the excited state will be: ( $R$ : Rydberg constant)
B According to Rydberg's formula, $\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{\mathrm{n}_{\mathrm{f}}^{2}}-\frac{1}{\mathrm{n}_{\mathrm{i}}^{2}}\right]$ $\mathrm{n}_{\mathrm{f}}=1$ $\mathrm{n}_{\mathrm{i}}=\mathrm{n}$ $\because \quad \frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{1^{2}}-\frac{1}{\mathrm{n}^{2}}\right]$ $\frac{1}{\lambda}=\mathrm{R}\left[1-\frac{1}{\mathrm{n}^{2}}\right]$ Multiplying by $\lambda$ on both side in equation (i), $1=\lambda \mathrm{R}\left[1-\frac{1}{\mathrm{n}^{2}}\right]$ $\frac{1}{\lambda \mathrm{R}}=1-\frac{1}{\mathrm{n}^{2}}$ $\frac{1}{\mathrm{n}^{2}}=1-\frac{1}{\lambda \mathrm{R}}$ $\frac{1}{\mathrm{n}^{2}}=\frac{\lambda \mathrm{R}-1}{\lambda \mathrm{R}}$ $\mathrm{n}=\sqrt{\frac{\lambda \mathrm{R}}{\lambda \mathrm{R}-1}}$
JEE Main-25.07.2022
ATOMS
145737
The total number of spectral lines observed when electron returns from the $6^{\text {th }}$ shell until the $2^{\text {nd }}$ shell in hydrogen atom is
1 15
2 10
3 20
4 25
Explanation:
B Given, $\mathrm{n}_{1}=2$ and $\mathrm{n}_{2}=6$ Total no. of spectral lines $=\frac{\left(\mathrm{n}_{2}-\mathrm{n}_{1}\right) \cdot\left(\mathrm{n}_{2}-\mathrm{n}_{1}+1\right)}{2}$ $=\frac{(6-2) \cdot(6-2+1)}{2}$ $=\frac{4 \times 5}{2}$ $=10$
TSEAMCET 18.07.2022
ATOMS
145738
The light emitted in the transition $n=3$ to $n=2$ (where $n$ is the principal quantum number of the state) in hydrogen is called $H_{\alpha}$-light find the maximum work function that a metal can have so that $H_{\alpha}$ light can emit photoelectrons from it.
1 $1.5 \mathrm{eV}$
2 $2.89 \mathrm{eV}$
3 $1.89 \mathrm{eV}$
4 $3.5 \mathrm{eV}$
Explanation:
C We have given that $\mathrm{n}_{1}=2$ $\mathrm{n}_{2}=3$ Energy possessed by $\mathrm{H}_{\alpha}$ light $=13.6\left(\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right)$ $=13.6\left[\frac{1}{(2)^{2}}-\frac{1}{(3)^{2}}\right]$ $=13.6\left[\frac{1}{4}-\frac{1}{9}\right]$ $=13.6 \times \frac{5}{36}$ $=\frac{68.0}{36}$ $=1.89 \mathrm{eV}$
TSEAMCET 31.07.2022
ATOMS
145739
An electron in a hydrogen atom makes a transition form $n=n_{1}$ to $n=n_{2}$ (where $n$ is a principal quantum number of a state). The time period of electron in the initial state is eight times than that of the final state, then which of the following statements is TRUE?
1 $\mathrm{n}_{1}=3 \mathrm{n}_{2}$
2 $\mathrm{n}_{1}=4 \mathrm{n}_{2}$
3 $\mathrm{n}_{1}=2 \mathrm{n}_{2}$
4 $\mathrm{n}_{1}=5 \mathrm{n}_{2}$
Explanation:
C The time period of electron in nth orbit $T_{n} \propto n^{3}$ Then, $\mathrm{T}_{1}=\mathrm{Kn}_{1}{ }^{3}$ $\mathrm{~T}_{2}=\mathrm{Kn}_{2}{ }^{3}$ Equation (ii) divided by equation (iii) we have $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\frac{\mathrm{Kn}_{1}^{3}}{\mathrm{Kn}_{2}^{3}}$ $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\left(\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}\right)^{3}$ According to question, $\mathrm{T}_{1}=8 \mathrm{~T}_{2}$ From equation (iv) $\frac{8 \mathrm{~T}_{2}}{\mathrm{~T}_{2}}=\left(\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}\right)^{3}$ $\left(\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}\right)^{3}=8$ $\left(\frac{n_{1}}{n_{2}}\right)^{3}=(2)^{3}$ $\frac{n_{1}}{n_{2}}=2$ $\mathrm{n}_{1}=2 \mathrm{n}_{2}$
145735
The energy levels of an hydrogen atom are shown below. The transition corresponding to emission of shortest wavelength is
1 $\mathrm{C}$
2 $\mathrm{D}$
3 $\mathrm{B}$
4 $\mathrm{A}$
Explanation:
B Since we know that, $\Delta \mathrm{E}=\frac{\mathrm{hc}}{\lambda}$ $\lambda \propto \frac{1}{\Delta \mathrm{E}}$ For shortest wavelength, energy gap should be maximum. So, correct choice is transition from $n=3$ to $n=1$.
JEE Main-06.04.2023
ATOMS
145736
Hydrogen atom from excited state comes to the ground by emitting a photon of wavelength $\lambda$. The value of principal quantum number ' $n$ ' of the excited state will be: ( $R$ : Rydberg constant)
B According to Rydberg's formula, $\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{\mathrm{n}_{\mathrm{f}}^{2}}-\frac{1}{\mathrm{n}_{\mathrm{i}}^{2}}\right]$ $\mathrm{n}_{\mathrm{f}}=1$ $\mathrm{n}_{\mathrm{i}}=\mathrm{n}$ $\because \quad \frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{1^{2}}-\frac{1}{\mathrm{n}^{2}}\right]$ $\frac{1}{\lambda}=\mathrm{R}\left[1-\frac{1}{\mathrm{n}^{2}}\right]$ Multiplying by $\lambda$ on both side in equation (i), $1=\lambda \mathrm{R}\left[1-\frac{1}{\mathrm{n}^{2}}\right]$ $\frac{1}{\lambda \mathrm{R}}=1-\frac{1}{\mathrm{n}^{2}}$ $\frac{1}{\mathrm{n}^{2}}=1-\frac{1}{\lambda \mathrm{R}}$ $\frac{1}{\mathrm{n}^{2}}=\frac{\lambda \mathrm{R}-1}{\lambda \mathrm{R}}$ $\mathrm{n}=\sqrt{\frac{\lambda \mathrm{R}}{\lambda \mathrm{R}-1}}$
JEE Main-25.07.2022
ATOMS
145737
The total number of spectral lines observed when electron returns from the $6^{\text {th }}$ shell until the $2^{\text {nd }}$ shell in hydrogen atom is
1 15
2 10
3 20
4 25
Explanation:
B Given, $\mathrm{n}_{1}=2$ and $\mathrm{n}_{2}=6$ Total no. of spectral lines $=\frac{\left(\mathrm{n}_{2}-\mathrm{n}_{1}\right) \cdot\left(\mathrm{n}_{2}-\mathrm{n}_{1}+1\right)}{2}$ $=\frac{(6-2) \cdot(6-2+1)}{2}$ $=\frac{4 \times 5}{2}$ $=10$
TSEAMCET 18.07.2022
ATOMS
145738
The light emitted in the transition $n=3$ to $n=2$ (where $n$ is the principal quantum number of the state) in hydrogen is called $H_{\alpha}$-light find the maximum work function that a metal can have so that $H_{\alpha}$ light can emit photoelectrons from it.
1 $1.5 \mathrm{eV}$
2 $2.89 \mathrm{eV}$
3 $1.89 \mathrm{eV}$
4 $3.5 \mathrm{eV}$
Explanation:
C We have given that $\mathrm{n}_{1}=2$ $\mathrm{n}_{2}=3$ Energy possessed by $\mathrm{H}_{\alpha}$ light $=13.6\left(\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right)$ $=13.6\left[\frac{1}{(2)^{2}}-\frac{1}{(3)^{2}}\right]$ $=13.6\left[\frac{1}{4}-\frac{1}{9}\right]$ $=13.6 \times \frac{5}{36}$ $=\frac{68.0}{36}$ $=1.89 \mathrm{eV}$
TSEAMCET 31.07.2022
ATOMS
145739
An electron in a hydrogen atom makes a transition form $n=n_{1}$ to $n=n_{2}$ (where $n$ is a principal quantum number of a state). The time period of electron in the initial state is eight times than that of the final state, then which of the following statements is TRUE?
1 $\mathrm{n}_{1}=3 \mathrm{n}_{2}$
2 $\mathrm{n}_{1}=4 \mathrm{n}_{2}$
3 $\mathrm{n}_{1}=2 \mathrm{n}_{2}$
4 $\mathrm{n}_{1}=5 \mathrm{n}_{2}$
Explanation:
C The time period of electron in nth orbit $T_{n} \propto n^{3}$ Then, $\mathrm{T}_{1}=\mathrm{Kn}_{1}{ }^{3}$ $\mathrm{~T}_{2}=\mathrm{Kn}_{2}{ }^{3}$ Equation (ii) divided by equation (iii) we have $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\frac{\mathrm{Kn}_{1}^{3}}{\mathrm{Kn}_{2}^{3}}$ $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\left(\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}\right)^{3}$ According to question, $\mathrm{T}_{1}=8 \mathrm{~T}_{2}$ From equation (iv) $\frac{8 \mathrm{~T}_{2}}{\mathrm{~T}_{2}}=\left(\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}\right)^{3}$ $\left(\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}\right)^{3}=8$ $\left(\frac{n_{1}}{n_{2}}\right)^{3}=(2)^{3}$ $\frac{n_{1}}{n_{2}}=2$ $\mathrm{n}_{1}=2 \mathrm{n}_{2}$
NEET Test Series from KOTA - 10 Papers In MS WORD
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ATOMS
145735
The energy levels of an hydrogen atom are shown below. The transition corresponding to emission of shortest wavelength is
1 $\mathrm{C}$
2 $\mathrm{D}$
3 $\mathrm{B}$
4 $\mathrm{A}$
Explanation:
B Since we know that, $\Delta \mathrm{E}=\frac{\mathrm{hc}}{\lambda}$ $\lambda \propto \frac{1}{\Delta \mathrm{E}}$ For shortest wavelength, energy gap should be maximum. So, correct choice is transition from $n=3$ to $n=1$.
JEE Main-06.04.2023
ATOMS
145736
Hydrogen atom from excited state comes to the ground by emitting a photon of wavelength $\lambda$. The value of principal quantum number ' $n$ ' of the excited state will be: ( $R$ : Rydberg constant)
B According to Rydberg's formula, $\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{\mathrm{n}_{\mathrm{f}}^{2}}-\frac{1}{\mathrm{n}_{\mathrm{i}}^{2}}\right]$ $\mathrm{n}_{\mathrm{f}}=1$ $\mathrm{n}_{\mathrm{i}}=\mathrm{n}$ $\because \quad \frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{1^{2}}-\frac{1}{\mathrm{n}^{2}}\right]$ $\frac{1}{\lambda}=\mathrm{R}\left[1-\frac{1}{\mathrm{n}^{2}}\right]$ Multiplying by $\lambda$ on both side in equation (i), $1=\lambda \mathrm{R}\left[1-\frac{1}{\mathrm{n}^{2}}\right]$ $\frac{1}{\lambda \mathrm{R}}=1-\frac{1}{\mathrm{n}^{2}}$ $\frac{1}{\mathrm{n}^{2}}=1-\frac{1}{\lambda \mathrm{R}}$ $\frac{1}{\mathrm{n}^{2}}=\frac{\lambda \mathrm{R}-1}{\lambda \mathrm{R}}$ $\mathrm{n}=\sqrt{\frac{\lambda \mathrm{R}}{\lambda \mathrm{R}-1}}$
JEE Main-25.07.2022
ATOMS
145737
The total number of spectral lines observed when electron returns from the $6^{\text {th }}$ shell until the $2^{\text {nd }}$ shell in hydrogen atom is
1 15
2 10
3 20
4 25
Explanation:
B Given, $\mathrm{n}_{1}=2$ and $\mathrm{n}_{2}=6$ Total no. of spectral lines $=\frac{\left(\mathrm{n}_{2}-\mathrm{n}_{1}\right) \cdot\left(\mathrm{n}_{2}-\mathrm{n}_{1}+1\right)}{2}$ $=\frac{(6-2) \cdot(6-2+1)}{2}$ $=\frac{4 \times 5}{2}$ $=10$
TSEAMCET 18.07.2022
ATOMS
145738
The light emitted in the transition $n=3$ to $n=2$ (where $n$ is the principal quantum number of the state) in hydrogen is called $H_{\alpha}$-light find the maximum work function that a metal can have so that $H_{\alpha}$ light can emit photoelectrons from it.
1 $1.5 \mathrm{eV}$
2 $2.89 \mathrm{eV}$
3 $1.89 \mathrm{eV}$
4 $3.5 \mathrm{eV}$
Explanation:
C We have given that $\mathrm{n}_{1}=2$ $\mathrm{n}_{2}=3$ Energy possessed by $\mathrm{H}_{\alpha}$ light $=13.6\left(\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right)$ $=13.6\left[\frac{1}{(2)^{2}}-\frac{1}{(3)^{2}}\right]$ $=13.6\left[\frac{1}{4}-\frac{1}{9}\right]$ $=13.6 \times \frac{5}{36}$ $=\frac{68.0}{36}$ $=1.89 \mathrm{eV}$
TSEAMCET 31.07.2022
ATOMS
145739
An electron in a hydrogen atom makes a transition form $n=n_{1}$ to $n=n_{2}$ (where $n$ is a principal quantum number of a state). The time period of electron in the initial state is eight times than that of the final state, then which of the following statements is TRUE?
1 $\mathrm{n}_{1}=3 \mathrm{n}_{2}$
2 $\mathrm{n}_{1}=4 \mathrm{n}_{2}$
3 $\mathrm{n}_{1}=2 \mathrm{n}_{2}$
4 $\mathrm{n}_{1}=5 \mathrm{n}_{2}$
Explanation:
C The time period of electron in nth orbit $T_{n} \propto n^{3}$ Then, $\mathrm{T}_{1}=\mathrm{Kn}_{1}{ }^{3}$ $\mathrm{~T}_{2}=\mathrm{Kn}_{2}{ }^{3}$ Equation (ii) divided by equation (iii) we have $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\frac{\mathrm{Kn}_{1}^{3}}{\mathrm{Kn}_{2}^{3}}$ $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\left(\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}\right)^{3}$ According to question, $\mathrm{T}_{1}=8 \mathrm{~T}_{2}$ From equation (iv) $\frac{8 \mathrm{~T}_{2}}{\mathrm{~T}_{2}}=\left(\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}\right)^{3}$ $\left(\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}\right)^{3}=8$ $\left(\frac{n_{1}}{n_{2}}\right)^{3}=(2)^{3}$ $\frac{n_{1}}{n_{2}}=2$ $\mathrm{n}_{1}=2 \mathrm{n}_{2}$
145735
The energy levels of an hydrogen atom are shown below. The transition corresponding to emission of shortest wavelength is
1 $\mathrm{C}$
2 $\mathrm{D}$
3 $\mathrm{B}$
4 $\mathrm{A}$
Explanation:
B Since we know that, $\Delta \mathrm{E}=\frac{\mathrm{hc}}{\lambda}$ $\lambda \propto \frac{1}{\Delta \mathrm{E}}$ For shortest wavelength, energy gap should be maximum. So, correct choice is transition from $n=3$ to $n=1$.
JEE Main-06.04.2023
ATOMS
145736
Hydrogen atom from excited state comes to the ground by emitting a photon of wavelength $\lambda$. The value of principal quantum number ' $n$ ' of the excited state will be: ( $R$ : Rydberg constant)
B According to Rydberg's formula, $\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{\mathrm{n}_{\mathrm{f}}^{2}}-\frac{1}{\mathrm{n}_{\mathrm{i}}^{2}}\right]$ $\mathrm{n}_{\mathrm{f}}=1$ $\mathrm{n}_{\mathrm{i}}=\mathrm{n}$ $\because \quad \frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{1^{2}}-\frac{1}{\mathrm{n}^{2}}\right]$ $\frac{1}{\lambda}=\mathrm{R}\left[1-\frac{1}{\mathrm{n}^{2}}\right]$ Multiplying by $\lambda$ on both side in equation (i), $1=\lambda \mathrm{R}\left[1-\frac{1}{\mathrm{n}^{2}}\right]$ $\frac{1}{\lambda \mathrm{R}}=1-\frac{1}{\mathrm{n}^{2}}$ $\frac{1}{\mathrm{n}^{2}}=1-\frac{1}{\lambda \mathrm{R}}$ $\frac{1}{\mathrm{n}^{2}}=\frac{\lambda \mathrm{R}-1}{\lambda \mathrm{R}}$ $\mathrm{n}=\sqrt{\frac{\lambda \mathrm{R}}{\lambda \mathrm{R}-1}}$
JEE Main-25.07.2022
ATOMS
145737
The total number of spectral lines observed when electron returns from the $6^{\text {th }}$ shell until the $2^{\text {nd }}$ shell in hydrogen atom is
1 15
2 10
3 20
4 25
Explanation:
B Given, $\mathrm{n}_{1}=2$ and $\mathrm{n}_{2}=6$ Total no. of spectral lines $=\frac{\left(\mathrm{n}_{2}-\mathrm{n}_{1}\right) \cdot\left(\mathrm{n}_{2}-\mathrm{n}_{1}+1\right)}{2}$ $=\frac{(6-2) \cdot(6-2+1)}{2}$ $=\frac{4 \times 5}{2}$ $=10$
TSEAMCET 18.07.2022
ATOMS
145738
The light emitted in the transition $n=3$ to $n=2$ (where $n$ is the principal quantum number of the state) in hydrogen is called $H_{\alpha}$-light find the maximum work function that a metal can have so that $H_{\alpha}$ light can emit photoelectrons from it.
1 $1.5 \mathrm{eV}$
2 $2.89 \mathrm{eV}$
3 $1.89 \mathrm{eV}$
4 $3.5 \mathrm{eV}$
Explanation:
C We have given that $\mathrm{n}_{1}=2$ $\mathrm{n}_{2}=3$ Energy possessed by $\mathrm{H}_{\alpha}$ light $=13.6\left(\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right)$ $=13.6\left[\frac{1}{(2)^{2}}-\frac{1}{(3)^{2}}\right]$ $=13.6\left[\frac{1}{4}-\frac{1}{9}\right]$ $=13.6 \times \frac{5}{36}$ $=\frac{68.0}{36}$ $=1.89 \mathrm{eV}$
TSEAMCET 31.07.2022
ATOMS
145739
An electron in a hydrogen atom makes a transition form $n=n_{1}$ to $n=n_{2}$ (where $n$ is a principal quantum number of a state). The time period of electron in the initial state is eight times than that of the final state, then which of the following statements is TRUE?
1 $\mathrm{n}_{1}=3 \mathrm{n}_{2}$
2 $\mathrm{n}_{1}=4 \mathrm{n}_{2}$
3 $\mathrm{n}_{1}=2 \mathrm{n}_{2}$
4 $\mathrm{n}_{1}=5 \mathrm{n}_{2}$
Explanation:
C The time period of electron in nth orbit $T_{n} \propto n^{3}$ Then, $\mathrm{T}_{1}=\mathrm{Kn}_{1}{ }^{3}$ $\mathrm{~T}_{2}=\mathrm{Kn}_{2}{ }^{3}$ Equation (ii) divided by equation (iii) we have $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\frac{\mathrm{Kn}_{1}^{3}}{\mathrm{Kn}_{2}^{3}}$ $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\left(\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}\right)^{3}$ According to question, $\mathrm{T}_{1}=8 \mathrm{~T}_{2}$ From equation (iv) $\frac{8 \mathrm{~T}_{2}}{\mathrm{~T}_{2}}=\left(\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}\right)^{3}$ $\left(\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}\right)^{3}=8$ $\left(\frac{n_{1}}{n_{2}}\right)^{3}=(2)^{3}$ $\frac{n_{1}}{n_{2}}=2$ $\mathrm{n}_{1}=2 \mathrm{n}_{2}$
