356412
The magnetic moment \(\left( {{m_{orb{\rm{ }}}}} \right)\) of a revolving electron around the nucleus varies with the principal quantum number \((n)\) as
1 \(m_{\text {orb }} \propto n^{2}\)
2 \(m_{\text {orb }} \propto \dfrac{1}{n^{2}}\)
3 \(m_{\text {orb }} \propto \dfrac{1}{n}\)
4 \(m_{\text {orb }} \propto n\)
Explanation:
For revolving electron magnetic moment \(m_{o r b}=\dfrac{e_{L}}{2 m_{e}}\) where \(L\) is angular momentum and \(m_{e}\) is mass of electron. By Bohr theory, \(L=n\left(\dfrac{h}{2 \pi}\right)\) \(\Rightarrow m_{o r b}=\dfrac{e_{n} h}{4 \pi m}\) \(\Rightarrow m_{o r b} \propto n\)
MHTCET - 2022
PHXII12:ATOMS
356413
The total energy of electron in the ground state of hydrogen atom is \( - 13.6\,eV\). The kinetic energy of an electron in the first excited state is
1 \(6.8\,eV\)
2 \(3.4\,eV\)
3 \(1.7\,eV\)
4 \(13.6\,eV\)
Explanation:
The energy of hydrogen atom when the electron revolves in \({n^{th}}\) orbit is given by \(T.E = \frac{{ - 13.6{Z^2}}}{{{n^2}}}eV\) \([Z = 1]\) For \(n = 2,\) \(E = \frac{{ - 13.6}}{{{2^2}}} = - 3.4\;eV\) So, kinetic energy of electron in the first excited state is \(KE = - T.E = - \left( { - 3.4} \right) = 3.4\;eV\)
PHXII12:ATOMS
356414
Angular momentum \((L)\) and radius \((r)\) of a hydrogen atom are related as:
356415
The ratio of speed of an electron in the ground state in Bohr’s first orbit of hydrogen atom to velocity of light (\(c\)) is (\(h = \) Planck’s constant, \({\varepsilon _0} = \) permitivity of free space, \(e = \)charge on electron)
1 \(\frac{{2{e^2}{\varepsilon _0}}}{{hc}}\)
2 \(\frac{{{e^3}}}{{2{\varepsilon _0}hc}}\)
3 \(\frac{{{e^2}}}{{2{\varepsilon _0}hc}}\)
4 \(\frac{{2{\varepsilon _0}hc}}{{{e^2}}}\)
Explanation:
The speed of electron in the \({n^{th}}\) state of hydrogen atom is \({v_n} = \left( {\frac{{{e^2}}}{{2h{\varepsilon _0}}}} \right)\frac{1}{n}m/s\) For ground state \((n = 1),{v_1} = \left( {\frac{{{e^2}}}{{2h{\varepsilon _0}}}} \right)m/s\) \(\therefore \frac{{{v_1}}}{c} = \frac{{{e^2}}}{{2{\varepsilon _0}hc}}\)
MHTCET - 2020
PHXII12:ATOMS
356416
Potential energy \((P{E_n})\) and kinetic energy \((K{E_n})\) of electron in \({n^{th}}\) orbit are related as:
356412
The magnetic moment \(\left( {{m_{orb{\rm{ }}}}} \right)\) of a revolving electron around the nucleus varies with the principal quantum number \((n)\) as
1 \(m_{\text {orb }} \propto n^{2}\)
2 \(m_{\text {orb }} \propto \dfrac{1}{n^{2}}\)
3 \(m_{\text {orb }} \propto \dfrac{1}{n}\)
4 \(m_{\text {orb }} \propto n\)
Explanation:
For revolving electron magnetic moment \(m_{o r b}=\dfrac{e_{L}}{2 m_{e}}\) where \(L\) is angular momentum and \(m_{e}\) is mass of electron. By Bohr theory, \(L=n\left(\dfrac{h}{2 \pi}\right)\) \(\Rightarrow m_{o r b}=\dfrac{e_{n} h}{4 \pi m}\) \(\Rightarrow m_{o r b} \propto n\)
MHTCET - 2022
PHXII12:ATOMS
356413
The total energy of electron in the ground state of hydrogen atom is \( - 13.6\,eV\). The kinetic energy of an electron in the first excited state is
1 \(6.8\,eV\)
2 \(3.4\,eV\)
3 \(1.7\,eV\)
4 \(13.6\,eV\)
Explanation:
The energy of hydrogen atom when the electron revolves in \({n^{th}}\) orbit is given by \(T.E = \frac{{ - 13.6{Z^2}}}{{{n^2}}}eV\) \([Z = 1]\) For \(n = 2,\) \(E = \frac{{ - 13.6}}{{{2^2}}} = - 3.4\;eV\) So, kinetic energy of electron in the first excited state is \(KE = - T.E = - \left( { - 3.4} \right) = 3.4\;eV\)
PHXII12:ATOMS
356414
Angular momentum \((L)\) and radius \((r)\) of a hydrogen atom are related as:
356415
The ratio of speed of an electron in the ground state in Bohr’s first orbit of hydrogen atom to velocity of light (\(c\)) is (\(h = \) Planck’s constant, \({\varepsilon _0} = \) permitivity of free space, \(e = \)charge on electron)
1 \(\frac{{2{e^2}{\varepsilon _0}}}{{hc}}\)
2 \(\frac{{{e^3}}}{{2{\varepsilon _0}hc}}\)
3 \(\frac{{{e^2}}}{{2{\varepsilon _0}hc}}\)
4 \(\frac{{2{\varepsilon _0}hc}}{{{e^2}}}\)
Explanation:
The speed of electron in the \({n^{th}}\) state of hydrogen atom is \({v_n} = \left( {\frac{{{e^2}}}{{2h{\varepsilon _0}}}} \right)\frac{1}{n}m/s\) For ground state \((n = 1),{v_1} = \left( {\frac{{{e^2}}}{{2h{\varepsilon _0}}}} \right)m/s\) \(\therefore \frac{{{v_1}}}{c} = \frac{{{e^2}}}{{2{\varepsilon _0}hc}}\)
MHTCET - 2020
PHXII12:ATOMS
356416
Potential energy \((P{E_n})\) and kinetic energy \((K{E_n})\) of electron in \({n^{th}}\) orbit are related as:
356412
The magnetic moment \(\left( {{m_{orb{\rm{ }}}}} \right)\) of a revolving electron around the nucleus varies with the principal quantum number \((n)\) as
1 \(m_{\text {orb }} \propto n^{2}\)
2 \(m_{\text {orb }} \propto \dfrac{1}{n^{2}}\)
3 \(m_{\text {orb }} \propto \dfrac{1}{n}\)
4 \(m_{\text {orb }} \propto n\)
Explanation:
For revolving electron magnetic moment \(m_{o r b}=\dfrac{e_{L}}{2 m_{e}}\) where \(L\) is angular momentum and \(m_{e}\) is mass of electron. By Bohr theory, \(L=n\left(\dfrac{h}{2 \pi}\right)\) \(\Rightarrow m_{o r b}=\dfrac{e_{n} h}{4 \pi m}\) \(\Rightarrow m_{o r b} \propto n\)
MHTCET - 2022
PHXII12:ATOMS
356413
The total energy of electron in the ground state of hydrogen atom is \( - 13.6\,eV\). The kinetic energy of an electron in the first excited state is
1 \(6.8\,eV\)
2 \(3.4\,eV\)
3 \(1.7\,eV\)
4 \(13.6\,eV\)
Explanation:
The energy of hydrogen atom when the electron revolves in \({n^{th}}\) orbit is given by \(T.E = \frac{{ - 13.6{Z^2}}}{{{n^2}}}eV\) \([Z = 1]\) For \(n = 2,\) \(E = \frac{{ - 13.6}}{{{2^2}}} = - 3.4\;eV\) So, kinetic energy of electron in the first excited state is \(KE = - T.E = - \left( { - 3.4} \right) = 3.4\;eV\)
PHXII12:ATOMS
356414
Angular momentum \((L)\) and radius \((r)\) of a hydrogen atom are related as:
356415
The ratio of speed of an electron in the ground state in Bohr’s first orbit of hydrogen atom to velocity of light (\(c\)) is (\(h = \) Planck’s constant, \({\varepsilon _0} = \) permitivity of free space, \(e = \)charge on electron)
1 \(\frac{{2{e^2}{\varepsilon _0}}}{{hc}}\)
2 \(\frac{{{e^3}}}{{2{\varepsilon _0}hc}}\)
3 \(\frac{{{e^2}}}{{2{\varepsilon _0}hc}}\)
4 \(\frac{{2{\varepsilon _0}hc}}{{{e^2}}}\)
Explanation:
The speed of electron in the \({n^{th}}\) state of hydrogen atom is \({v_n} = \left( {\frac{{{e^2}}}{{2h{\varepsilon _0}}}} \right)\frac{1}{n}m/s\) For ground state \((n = 1),{v_1} = \left( {\frac{{{e^2}}}{{2h{\varepsilon _0}}}} \right)m/s\) \(\therefore \frac{{{v_1}}}{c} = \frac{{{e^2}}}{{2{\varepsilon _0}hc}}\)
MHTCET - 2020
PHXII12:ATOMS
356416
Potential energy \((P{E_n})\) and kinetic energy \((K{E_n})\) of electron in \({n^{th}}\) orbit are related as:
356412
The magnetic moment \(\left( {{m_{orb{\rm{ }}}}} \right)\) of a revolving electron around the nucleus varies with the principal quantum number \((n)\) as
1 \(m_{\text {orb }} \propto n^{2}\)
2 \(m_{\text {orb }} \propto \dfrac{1}{n^{2}}\)
3 \(m_{\text {orb }} \propto \dfrac{1}{n}\)
4 \(m_{\text {orb }} \propto n\)
Explanation:
For revolving electron magnetic moment \(m_{o r b}=\dfrac{e_{L}}{2 m_{e}}\) where \(L\) is angular momentum and \(m_{e}\) is mass of electron. By Bohr theory, \(L=n\left(\dfrac{h}{2 \pi}\right)\) \(\Rightarrow m_{o r b}=\dfrac{e_{n} h}{4 \pi m}\) \(\Rightarrow m_{o r b} \propto n\)
MHTCET - 2022
PHXII12:ATOMS
356413
The total energy of electron in the ground state of hydrogen atom is \( - 13.6\,eV\). The kinetic energy of an electron in the first excited state is
1 \(6.8\,eV\)
2 \(3.4\,eV\)
3 \(1.7\,eV\)
4 \(13.6\,eV\)
Explanation:
The energy of hydrogen atom when the electron revolves in \({n^{th}}\) orbit is given by \(T.E = \frac{{ - 13.6{Z^2}}}{{{n^2}}}eV\) \([Z = 1]\) For \(n = 2,\) \(E = \frac{{ - 13.6}}{{{2^2}}} = - 3.4\;eV\) So, kinetic energy of electron in the first excited state is \(KE = - T.E = - \left( { - 3.4} \right) = 3.4\;eV\)
PHXII12:ATOMS
356414
Angular momentum \((L)\) and radius \((r)\) of a hydrogen atom are related as:
356415
The ratio of speed of an electron in the ground state in Bohr’s first orbit of hydrogen atom to velocity of light (\(c\)) is (\(h = \) Planck’s constant, \({\varepsilon _0} = \) permitivity of free space, \(e = \)charge on electron)
1 \(\frac{{2{e^2}{\varepsilon _0}}}{{hc}}\)
2 \(\frac{{{e^3}}}{{2{\varepsilon _0}hc}}\)
3 \(\frac{{{e^2}}}{{2{\varepsilon _0}hc}}\)
4 \(\frac{{2{\varepsilon _0}hc}}{{{e^2}}}\)
Explanation:
The speed of electron in the \({n^{th}}\) state of hydrogen atom is \({v_n} = \left( {\frac{{{e^2}}}{{2h{\varepsilon _0}}}} \right)\frac{1}{n}m/s\) For ground state \((n = 1),{v_1} = \left( {\frac{{{e^2}}}{{2h{\varepsilon _0}}}} \right)m/s\) \(\therefore \frac{{{v_1}}}{c} = \frac{{{e^2}}}{{2{\varepsilon _0}hc}}\)
MHTCET - 2020
PHXII12:ATOMS
356416
Potential energy \((P{E_n})\) and kinetic energy \((K{E_n})\) of electron in \({n^{th}}\) orbit are related as:
356412
The magnetic moment \(\left( {{m_{orb{\rm{ }}}}} \right)\) of a revolving electron around the nucleus varies with the principal quantum number \((n)\) as
1 \(m_{\text {orb }} \propto n^{2}\)
2 \(m_{\text {orb }} \propto \dfrac{1}{n^{2}}\)
3 \(m_{\text {orb }} \propto \dfrac{1}{n}\)
4 \(m_{\text {orb }} \propto n\)
Explanation:
For revolving electron magnetic moment \(m_{o r b}=\dfrac{e_{L}}{2 m_{e}}\) where \(L\) is angular momentum and \(m_{e}\) is mass of electron. By Bohr theory, \(L=n\left(\dfrac{h}{2 \pi}\right)\) \(\Rightarrow m_{o r b}=\dfrac{e_{n} h}{4 \pi m}\) \(\Rightarrow m_{o r b} \propto n\)
MHTCET - 2022
PHXII12:ATOMS
356413
The total energy of electron in the ground state of hydrogen atom is \( - 13.6\,eV\). The kinetic energy of an electron in the first excited state is
1 \(6.8\,eV\)
2 \(3.4\,eV\)
3 \(1.7\,eV\)
4 \(13.6\,eV\)
Explanation:
The energy of hydrogen atom when the electron revolves in \({n^{th}}\) orbit is given by \(T.E = \frac{{ - 13.6{Z^2}}}{{{n^2}}}eV\) \([Z = 1]\) For \(n = 2,\) \(E = \frac{{ - 13.6}}{{{2^2}}} = - 3.4\;eV\) So, kinetic energy of electron in the first excited state is \(KE = - T.E = - \left( { - 3.4} \right) = 3.4\;eV\)
PHXII12:ATOMS
356414
Angular momentum \((L)\) and radius \((r)\) of a hydrogen atom are related as:
356415
The ratio of speed of an electron in the ground state in Bohr’s first orbit of hydrogen atom to velocity of light (\(c\)) is (\(h = \) Planck’s constant, \({\varepsilon _0} = \) permitivity of free space, \(e = \)charge on electron)
1 \(\frac{{2{e^2}{\varepsilon _0}}}{{hc}}\)
2 \(\frac{{{e^3}}}{{2{\varepsilon _0}hc}}\)
3 \(\frac{{{e^2}}}{{2{\varepsilon _0}hc}}\)
4 \(\frac{{2{\varepsilon _0}hc}}{{{e^2}}}\)
Explanation:
The speed of electron in the \({n^{th}}\) state of hydrogen atom is \({v_n} = \left( {\frac{{{e^2}}}{{2h{\varepsilon _0}}}} \right)\frac{1}{n}m/s\) For ground state \((n = 1),{v_1} = \left( {\frac{{{e^2}}}{{2h{\varepsilon _0}}}} \right)m/s\) \(\therefore \frac{{{v_1}}}{c} = \frac{{{e^2}}}{{2{\varepsilon _0}hc}}\)
MHTCET - 2020
PHXII12:ATOMS
356416
Potential energy \((P{E_n})\) and kinetic energy \((K{E_n})\) of electron in \({n^{th}}\) orbit are related as:
