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PHXII12:ATOMS

356504 The ratio of ionization energy of Bohr's hydrogen atom and Bohr's hydrogen like Lithium atom is

1 \({1: 1}\)

2 \({1: 3}\)

3 \({1: 9}\)

4 \({1: 27}\)

PHXII12:ATOMS

356505 A small particle of mass \(m\) moves in such a way that its potential energy \(U=\dfrac{1}{2} m \omega^{2} r^{2}\), where \(\omega\) is constant and \(r\) is the distance of the particle from origin. Assuming Bohr's quantization of momentum and circular orbit, the radius of \(n^{\text {th }}\) orbit will be proportional to

1 \(\sqrt{n}\)

2 \(n^{2}\)

3 \(n\)

4 \(\dfrac{1}{n}\)

JEE - 2023

PHXII12:ATOMS

356506 The angular momentum of an electron in the \({2^{nd}}\) excited state of Helium ion \((H{e^ + })\) is

1 \(\frac{h}{{2\pi }}\)

2 \(\frac{{2h}}{{2\pi }}\)

3 \(\frac{{3h}}{{2\pi }}\)

4 \(\frac{{4h}}{{2\pi }}\)

PHXII12:ATOMS

356507 A hydrogen-like atom of atomic number \({Z}\) is in an excited state of quantum number \({2 n}\). It can emit a maximum energy photon of 204 \(eV\) . If it makes a transition to quantum state \({n}\), a photon of energy 40.8 \(eV\) is emitted. Find the minimum energy that can be emitted by this atom during de-excitation. Ground state energy of hydrogen atom is \(-\)13.6 \(eV\) .

1 \(15.62\,eV\)

2 \(10.58\,eV\)

3 \(20.16\,eV\)

4 \(24.58\,eV\)

Explanation:

The energy levels of a hydrogen-like atom are governed by
\(\begin{equation*}E_{n}=-\dfrac{Z^{2} R c h}{n^{2}}=-(13.6 {eV}) \dfrac{Z^{2}}{n^{2}} \tag{1}\end{equation*}\)
The energy of emitted photon corresponding to transition
\(n=n_{2}\) to \(n=n_{1}\) with \(n_{2}>n_{1}\) is given by
\({E_{{n_2} \to {n_1}}} = - (13.6eV){Z^2}\left( {\frac{1}{{n_2^2}} - \frac{1}{{n_1^2}}} \right)\,\,\,\,\,\,\,(2)\)
If \(n_{2}=2 n\), the energy of the emitted photon will be maximum for transition \(n_{2}=2 n\) to \(n_{1}=1\). Thus
\(E_{\max }=-(13.6 {eV}) Z^{2}\left(\dfrac{1}{(2 n)^{2}}-\dfrac{1}{1^{2}}\right)\)
\(\Rightarrow E_{\max }=-(13.6 {eV}) Z^{2}\left(\dfrac{1}{4 n^{2}}-1\right)\)
Given that \(E_{\max }=204 {eV}\), we use
\(204eV = - (13.6eV){Z^2}\left( {\frac{1}{{4{n^2}}} - 1} \right)\,\,\,\,\,\,\,\,\,\,(3)\)
for the transition \(n_{2}=2 n\) to \(n_{1}=n\), we have
\(E_{2 n \rightarrow n}=-(13.6 {eV}) Z^{2}\left(\dfrac{1}{4 n^{2}}-\dfrac{1}{n^{2}}\right)\)
Given that \(E_{2 n \rightarrow n}=40.8 {eV}\), we use
\(40.8eV = - (13.6eV){Z^2}\left( {\frac{1}{{4{n^2}}} - \frac{1}{{{n^2}}}} \right)\,\,\,\,\,\,\,\,(4)\)
Dividing (3) by (4), we get
\(\dfrac{204}{40.8}=\dfrac{\left(\dfrac{1}{4 n^{2}}-1\right)}{\left(\dfrac{1}{4 n^{2}}-\dfrac{1}{n^{2}}\right)} \Rightarrow 5=\dfrac{1-4 n^{2}}{1-4}\)
Which gives \(4 n^{2}=16\) or \(n=2\). Using this value of \(n\) in either (3) or (4) we get \(Z^{2}=16\) or \(Z=4\).
The ground state energy of the atom corresponds to state \(n=1\). Putting \(n=1\) and \(Z=4\) in Eq. (1), the ground state energy of the atom is
\(E_{1}=-(13.6 {eV}) \times \dfrac{(4)^{2}}{(1)^{2}}=-217.6 {eV}\)
Since \(n=2\), the given excited atom has \(n_{2}=2 n=4\).
It will emit a photon of minimum energy for a transition \(n_{2}=4\,\,to\)
\(n_{1}=3\). Using these values in (2), we have
\(E_{\min }=(-13.6 {eV})(4)^{2}\left(\dfrac{1}{4^{2}}-\dfrac{1}{3^{2}}\right)=10.58 {eV}\)

PHXII12:ATOMS

356504 The ratio of ionization energy of Bohr's hydrogen atom and Bohr's hydrogen like Lithium atom is

1 \({1: 1}\)

2 \({1: 3}\)

3 \({1: 9}\)

4 \({1: 27}\)

PHXII12:ATOMS

356505 A small particle of mass \(m\) moves in such a way that its potential energy \(U=\dfrac{1}{2} m \omega^{2} r^{2}\), where \(\omega\) is constant and \(r\) is the distance of the particle from origin. Assuming Bohr's quantization of momentum and circular orbit, the radius of \(n^{\text {th }}\) orbit will be proportional to

1 \(\sqrt{n}\)

2 \(n^{2}\)

3 \(n\)

4 \(\dfrac{1}{n}\)

JEE - 2023

PHXII12:ATOMS

356506 The angular momentum of an electron in the \({2^{nd}}\) excited state of Helium ion \((H{e^ + })\) is

1 \(\frac{h}{{2\pi }}\)

2 \(\frac{{2h}}{{2\pi }}\)

3 \(\frac{{3h}}{{2\pi }}\)

4 \(\frac{{4h}}{{2\pi }}\)

PHXII12:ATOMS

356507 A hydrogen-like atom of atomic number \({Z}\) is in an excited state of quantum number \({2 n}\). It can emit a maximum energy photon of 204 \(eV\) . If it makes a transition to quantum state \({n}\), a photon of energy 40.8 \(eV\) is emitted. Find the minimum energy that can be emitted by this atom during de-excitation. Ground state energy of hydrogen atom is \(-\)13.6 \(eV\) .

1 \(15.62\,eV\)

2 \(10.58\,eV\)

3 \(20.16\,eV\)

4 \(24.58\,eV\)

Explanation:

The energy levels of a hydrogen-like atom are governed by
\(\begin{equation*}E_{n}=-\dfrac{Z^{2} R c h}{n^{2}}=-(13.6 {eV}) \dfrac{Z^{2}}{n^{2}} \tag{1}\end{equation*}\)
The energy of emitted photon corresponding to transition
\(n=n_{2}\) to \(n=n_{1}\) with \(n_{2}>n_{1}\) is given by
\({E_{{n_2} \to {n_1}}} = - (13.6eV){Z^2}\left( {\frac{1}{{n_2^2}} - \frac{1}{{n_1^2}}} \right)\,\,\,\,\,\,\,(2)\)
If \(n_{2}=2 n\), the energy of the emitted photon will be maximum for transition \(n_{2}=2 n\) to \(n_{1}=1\). Thus
\(E_{\max }=-(13.6 {eV}) Z^{2}\left(\dfrac{1}{(2 n)^{2}}-\dfrac{1}{1^{2}}\right)\)
\(\Rightarrow E_{\max }=-(13.6 {eV}) Z^{2}\left(\dfrac{1}{4 n^{2}}-1\right)\)
Given that \(E_{\max }=204 {eV}\), we use
\(204eV = - (13.6eV){Z^2}\left( {\frac{1}{{4{n^2}}} - 1} \right)\,\,\,\,\,\,\,\,\,\,(3)\)
for the transition \(n_{2}=2 n\) to \(n_{1}=n\), we have
\(E_{2 n \rightarrow n}=-(13.6 {eV}) Z^{2}\left(\dfrac{1}{4 n^{2}}-\dfrac{1}{n^{2}}\right)\)
Given that \(E_{2 n \rightarrow n}=40.8 {eV}\), we use
\(40.8eV = - (13.6eV){Z^2}\left( {\frac{1}{{4{n^2}}} - \frac{1}{{{n^2}}}} \right)\,\,\,\,\,\,\,\,(4)\)
Dividing (3) by (4), we get
\(\dfrac{204}{40.8}=\dfrac{\left(\dfrac{1}{4 n^{2}}-1\right)}{\left(\dfrac{1}{4 n^{2}}-\dfrac{1}{n^{2}}\right)} \Rightarrow 5=\dfrac{1-4 n^{2}}{1-4}\)
Which gives \(4 n^{2}=16\) or \(n=2\). Using this value of \(n\) in either (3) or (4) we get \(Z^{2}=16\) or \(Z=4\).
The ground state energy of the atom corresponds to state \(n=1\). Putting \(n=1\) and \(Z=4\) in Eq. (1), the ground state energy of the atom is
\(E_{1}=-(13.6 {eV}) \times \dfrac{(4)^{2}}{(1)^{2}}=-217.6 {eV}\)
Since \(n=2\), the given excited atom has \(n_{2}=2 n=4\).
It will emit a photon of minimum energy for a transition \(n_{2}=4\,\,to\)
\(n_{1}=3\). Using these values in (2), we have
\(E_{\min }=(-13.6 {eV})(4)^{2}\left(\dfrac{1}{4^{2}}-\dfrac{1}{3^{2}}\right)=10.58 {eV}\)

PHXII12:ATOMS

356504 The ratio of ionization energy of Bohr's hydrogen atom and Bohr's hydrogen like Lithium atom is

1 \({1: 1}\)

2 \({1: 3}\)

3 \({1: 9}\)

4 \({1: 27}\)

PHXII12:ATOMS

356505 A small particle of mass \(m\) moves in such a way that its potential energy \(U=\dfrac{1}{2} m \omega^{2} r^{2}\), where \(\omega\) is constant and \(r\) is the distance of the particle from origin. Assuming Bohr's quantization of momentum and circular orbit, the radius of \(n^{\text {th }}\) orbit will be proportional to

1 \(\sqrt{n}\)

2 \(n^{2}\)

3 \(n\)

4 \(\dfrac{1}{n}\)

JEE - 2023

PHXII12:ATOMS

356506 The angular momentum of an electron in the \({2^{nd}}\) excited state of Helium ion \((H{e^ + })\) is

1 \(\frac{h}{{2\pi }}\)

2 \(\frac{{2h}}{{2\pi }}\)

3 \(\frac{{3h}}{{2\pi }}\)

4 \(\frac{{4h}}{{2\pi }}\)

PHXII12:ATOMS

356507 A hydrogen-like atom of atomic number \({Z}\) is in an excited state of quantum number \({2 n}\). It can emit a maximum energy photon of 204 \(eV\) . If it makes a transition to quantum state \({n}\), a photon of energy 40.8 \(eV\) is emitted. Find the minimum energy that can be emitted by this atom during de-excitation. Ground state energy of hydrogen atom is \(-\)13.6 \(eV\) .

1 \(15.62\,eV\)

2 \(10.58\,eV\)

3 \(20.16\,eV\)

4 \(24.58\,eV\)

Explanation:

The energy levels of a hydrogen-like atom are governed by
\(\begin{equation*}E_{n}=-\dfrac{Z^{2} R c h}{n^{2}}=-(13.6 {eV}) \dfrac{Z^{2}}{n^{2}} \tag{1}\end{equation*}\)
The energy of emitted photon corresponding to transition
\(n=n_{2}\) to \(n=n_{1}\) with \(n_{2}>n_{1}\) is given by
\({E_{{n_2} \to {n_1}}} = - (13.6eV){Z^2}\left( {\frac{1}{{n_2^2}} - \frac{1}{{n_1^2}}} \right)\,\,\,\,\,\,\,(2)\)
If \(n_{2}=2 n\), the energy of the emitted photon will be maximum for transition \(n_{2}=2 n\) to \(n_{1}=1\). Thus
\(E_{\max }=-(13.6 {eV}) Z^{2}\left(\dfrac{1}{(2 n)^{2}}-\dfrac{1}{1^{2}}\right)\)
\(\Rightarrow E_{\max }=-(13.6 {eV}) Z^{2}\left(\dfrac{1}{4 n^{2}}-1\right)\)
Given that \(E_{\max }=204 {eV}\), we use
\(204eV = - (13.6eV){Z^2}\left( {\frac{1}{{4{n^2}}} - 1} \right)\,\,\,\,\,\,\,\,\,\,(3)\)
for the transition \(n_{2}=2 n\) to \(n_{1}=n\), we have
\(E_{2 n \rightarrow n}=-(13.6 {eV}) Z^{2}\left(\dfrac{1}{4 n^{2}}-\dfrac{1}{n^{2}}\right)\)
Given that \(E_{2 n \rightarrow n}=40.8 {eV}\), we use
\(40.8eV = - (13.6eV){Z^2}\left( {\frac{1}{{4{n^2}}} - \frac{1}{{{n^2}}}} \right)\,\,\,\,\,\,\,\,(4)\)
Dividing (3) by (4), we get
\(\dfrac{204}{40.8}=\dfrac{\left(\dfrac{1}{4 n^{2}}-1\right)}{\left(\dfrac{1}{4 n^{2}}-\dfrac{1}{n^{2}}\right)} \Rightarrow 5=\dfrac{1-4 n^{2}}{1-4}\)
Which gives \(4 n^{2}=16\) or \(n=2\). Using this value of \(n\) in either (3) or (4) we get \(Z^{2}=16\) or \(Z=4\).
The ground state energy of the atom corresponds to state \(n=1\). Putting \(n=1\) and \(Z=4\) in Eq. (1), the ground state energy of the atom is
\(E_{1}=-(13.6 {eV}) \times \dfrac{(4)^{2}}{(1)^{2}}=-217.6 {eV}\)
Since \(n=2\), the given excited atom has \(n_{2}=2 n=4\).
It will emit a photon of minimum energy for a transition \(n_{2}=4\,\,to\)
\(n_{1}=3\). Using these values in (2), we have
\(E_{\min }=(-13.6 {eV})(4)^{2}\left(\dfrac{1}{4^{2}}-\dfrac{1}{3^{2}}\right)=10.58 {eV}\)

PHXII12:ATOMS

356504 The ratio of ionization energy of Bohr's hydrogen atom and Bohr's hydrogen like Lithium atom is

1 \({1: 1}\)

2 \({1: 3}\)

3 \({1: 9}\)

4 \({1: 27}\)

PHXII12:ATOMS

356505 A small particle of mass \(m\) moves in such a way that its potential energy \(U=\dfrac{1}{2} m \omega^{2} r^{2}\), where \(\omega\) is constant and \(r\) is the distance of the particle from origin. Assuming Bohr's quantization of momentum and circular orbit, the radius of \(n^{\text {th }}\) orbit will be proportional to

1 \(\sqrt{n}\)

2 \(n^{2}\)

3 \(n\)

4 \(\dfrac{1}{n}\)

JEE - 2023

PHXII12:ATOMS

356506 The angular momentum of an electron in the \({2^{nd}}\) excited state of Helium ion \((H{e^ + })\) is

1 \(\frac{h}{{2\pi }}\)

2 \(\frac{{2h}}{{2\pi }}\)

3 \(\frac{{3h}}{{2\pi }}\)

4 \(\frac{{4h}}{{2\pi }}\)

PHXII12:ATOMS

356507 A hydrogen-like atom of atomic number \({Z}\) is in an excited state of quantum number \({2 n}\). It can emit a maximum energy photon of 204 \(eV\) . If it makes a transition to quantum state \({n}\), a photon of energy 40.8 \(eV\) is emitted. Find the minimum energy that can be emitted by this atom during de-excitation. Ground state energy of hydrogen atom is \(-\)13.6 \(eV\) .

1 \(15.62\,eV\)

2 \(10.58\,eV\)

3 \(20.16\,eV\)

4 \(24.58\,eV\)

Explanation:

The energy levels of a hydrogen-like atom are governed by
\(\begin{equation*}E_{n}=-\dfrac{Z^{2} R c h}{n^{2}}=-(13.6 {eV}) \dfrac{Z^{2}}{n^{2}} \tag{1}\end{equation*}\)
The energy of emitted photon corresponding to transition
\(n=n_{2}\) to \(n=n_{1}\) with \(n_{2}>n_{1}\) is given by
\({E_{{n_2} \to {n_1}}} = - (13.6eV){Z^2}\left( {\frac{1}{{n_2^2}} - \frac{1}{{n_1^2}}} \right)\,\,\,\,\,\,\,(2)\)
If \(n_{2}=2 n\), the energy of the emitted photon will be maximum for transition \(n_{2}=2 n\) to \(n_{1}=1\). Thus
\(E_{\max }=-(13.6 {eV}) Z^{2}\left(\dfrac{1}{(2 n)^{2}}-\dfrac{1}{1^{2}}\right)\)
\(\Rightarrow E_{\max }=-(13.6 {eV}) Z^{2}\left(\dfrac{1}{4 n^{2}}-1\right)\)
Given that \(E_{\max }=204 {eV}\), we use
\(204eV = - (13.6eV){Z^2}\left( {\frac{1}{{4{n^2}}} - 1} \right)\,\,\,\,\,\,\,\,\,\,(3)\)
for the transition \(n_{2}=2 n\) to \(n_{1}=n\), we have
\(E_{2 n \rightarrow n}=-(13.6 {eV}) Z^{2}\left(\dfrac{1}{4 n^{2}}-\dfrac{1}{n^{2}}\right)\)
Given that \(E_{2 n \rightarrow n}=40.8 {eV}\), we use
\(40.8eV = - (13.6eV){Z^2}\left( {\frac{1}{{4{n^2}}} - \frac{1}{{{n^2}}}} \right)\,\,\,\,\,\,\,\,(4)\)
Dividing (3) by (4), we get
\(\dfrac{204}{40.8}=\dfrac{\left(\dfrac{1}{4 n^{2}}-1\right)}{\left(\dfrac{1}{4 n^{2}}-\dfrac{1}{n^{2}}\right)} \Rightarrow 5=\dfrac{1-4 n^{2}}{1-4}\)
Which gives \(4 n^{2}=16\) or \(n=2\). Using this value of \(n\) in either (3) or (4) we get \(Z^{2}=16\) or \(Z=4\).
The ground state energy of the atom corresponds to state \(n=1\). Putting \(n=1\) and \(Z=4\) in Eq. (1), the ground state energy of the atom is
\(E_{1}=-(13.6 {eV}) \times \dfrac{(4)^{2}}{(1)^{2}}=-217.6 {eV}\)
Since \(n=2\), the given excited atom has \(n_{2}=2 n=4\).
It will emit a photon of minimum energy for a transition \(n_{2}=4\,\,to\)
\(n_{1}=3\). Using these values in (2), we have
\(E_{\min }=(-13.6 {eV})(4)^{2}\left(\dfrac{1}{4^{2}}-\dfrac{1}{3^{2}}\right)=10.58 {eV}\)

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