QBManager

PHXII12:ATOMS

356430 The effective current due to the rotation of electron in atom depends on $Z$ and $n$ as

1

\frac{Z}{n^{2}}

2

\frac{Z}{n}

3

\frac{Z^{2}}{n^{3}}

4

\frac{Z^{2}}{n}

Explanation:

$i = \frac{e}{T} = \frac{e v_{n}}{2 π r_{n}} = \frac{m Z^{2} e^{5}}{4 ε_{0}^{2} h^{3} n^{3}}$

PHXII12:ATOMS

356431 The ratio of magnetic dipole moment to angular momentum in a hydrogen atom is

1

\frac{e}{m}

2

\frac{e}{2 m}

3

\frac{e}{3 m}

4

\frac{2 e}{m}

Explanation:

Magnetic moment
$μ = i A = \frac{q}{T} π r^{2} = \frac{q ω r^{2}}{2}$
Angular momentum
$L = m r^{2} ω$
$\frac{μ}{L} = \frac{q}{2 m} = \frac{e}{2 m}$

PHXII12:ATOMS

356432 Let the $P E$ of hydrogen atom in the ground state be zero. Then its total energy in the first excited state will be

1

23.8 e V

2

27.2 e V

3

10.2 e V

4

12.6 e V

Explanation:

When the $P$ . $E$ is taken as zero (Old-Frame) at infinity then the values of $P$ . $E$ , $K$ . $E$ & $T$ . $E$ for $H$ -atom are shown as follows
$P . E = 2 T . E & K . E = - T . E$
supporting img
$T . E = \frac{- 13.6 e V}{n^{2}}$

The above relations between $P$ . $E$ , $T$ . $E$ & $K$ . $E$ are not valid in the new frame.
Note: If we change the reference point for $P$ . $E$ then the absolute $K$ . $E$ and difference in $T$ . $E$ must remain same.

PHXII12:ATOMS

356433 Statement A :
The total energy of an electron revolving in any stationary orbit is negative.
Statement B :
Kinetic energy of revolving electron can be $(- v e)$ also.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both Statements are correct.

4 Both Statements are incorrect.

Explanation:

Energy can have positive or negative values. In the atom the electrons are bound. The negative energy of the revolving electrons does not allow the electrons to leave the atom. Kinetic energy of the electron is always $(+) v e$ . So option (1) is correct.

PHXII12:ATOMS

356430 The effective current due to the rotation of electron in atom depends on $Z$ and $n$ as

1

\frac{Z}{n^{2}}

2

\frac{Z}{n}

3

\frac{Z^{2}}{n^{3}}

4

\frac{Z^{2}}{n}

Explanation:

$i = \frac{e}{T} = \frac{e v_{n}}{2 π r_{n}} = \frac{m Z^{2} e^{5}}{4 ε_{0}^{2} h^{3} n^{3}}$

PHXII12:ATOMS

356431 The ratio of magnetic dipole moment to angular momentum in a hydrogen atom is

1

\frac{e}{m}

2

\frac{e}{2 m}

3

\frac{e}{3 m}

4

\frac{2 e}{m}

Explanation:

Magnetic moment
$μ = i A = \frac{q}{T} π r^{2} = \frac{q ω r^{2}}{2}$
Angular momentum
$L = m r^{2} ω$
$\frac{μ}{L} = \frac{q}{2 m} = \frac{e}{2 m}$

PHXII12:ATOMS

356432 Let the $P E$ of hydrogen atom in the ground state be zero. Then its total energy in the first excited state will be

1

23.8 e V

2

27.2 e V

3

10.2 e V

4

12.6 e V

Explanation:

When the $P$ . $E$ is taken as zero (Old-Frame) at infinity then the values of $P$ . $E$ , $K$ . $E$ & $T$ . $E$ for $H$ -atom are shown as follows
$P . E = 2 T . E & K . E = - T . E$
supporting img
$T . E = \frac{- 13.6 e V}{n^{2}}$

The above relations between $P$ . $E$ , $T$ . $E$ & $K$ . $E$ are not valid in the new frame.
Note: If we change the reference point for $P$ . $E$ then the absolute $K$ . $E$ and difference in $T$ . $E$ must remain same.

PHXII12:ATOMS

356433 Statement A :
The total energy of an electron revolving in any stationary orbit is negative.
Statement B :
Kinetic energy of revolving electron can be $(- v e)$ also.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both Statements are correct.

4 Both Statements are incorrect.

Explanation:

Energy can have positive or negative values. In the atom the electrons are bound. The negative energy of the revolving electrons does not allow the electrons to leave the atom. Kinetic energy of the electron is always $(+) v e$ . So option (1) is correct.

PHXII12:ATOMS

356430 The effective current due to the rotation of electron in atom depends on $Z$ and $n$ as

1

\frac{Z}{n^{2}}

2

\frac{Z}{n}

3

\frac{Z^{2}}{n^{3}}

4

\frac{Z^{2}}{n}

Explanation:

$i = \frac{e}{T} = \frac{e v_{n}}{2 π r_{n}} = \frac{m Z^{2} e^{5}}{4 ε_{0}^{2} h^{3} n^{3}}$

PHXII12:ATOMS

356431 The ratio of magnetic dipole moment to angular momentum in a hydrogen atom is

1

\frac{e}{m}

2

\frac{e}{2 m}

3

\frac{e}{3 m}

4

\frac{2 e}{m}

Explanation:

Magnetic moment
$μ = i A = \frac{q}{T} π r^{2} = \frac{q ω r^{2}}{2}$
Angular momentum
$L = m r^{2} ω$
$\frac{μ}{L} = \frac{q}{2 m} = \frac{e}{2 m}$

PHXII12:ATOMS

356432 Let the $P E$ of hydrogen atom in the ground state be zero. Then its total energy in the first excited state will be

1

23.8 e V

2

27.2 e V

3

10.2 e V

4

12.6 e V

Explanation:

When the $P$ . $E$ is taken as zero (Old-Frame) at infinity then the values of $P$ . $E$ , $K$ . $E$ & $T$ . $E$ for $H$ -atom are shown as follows
$P . E = 2 T . E & K . E = - T . E$
supporting img
$T . E = \frac{- 13.6 e V}{n^{2}}$

The above relations between $P$ . $E$ , $T$ . $E$ & $K$ . $E$ are not valid in the new frame.
Note: If we change the reference point for $P$ . $E$ then the absolute $K$ . $E$ and difference in $T$ . $E$ must remain same.

PHXII12:ATOMS

356433 Statement A :
The total energy of an electron revolving in any stationary orbit is negative.
Statement B :
Kinetic energy of revolving electron can be $(- v e)$ also.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both Statements are correct.

4 Both Statements are incorrect.

Explanation:

Energy can have positive or negative values. In the atom the electrons are bound. The negative energy of the revolving electrons does not allow the electrons to leave the atom. Kinetic energy of the electron is always $(+) v e$ . So option (1) is correct.

PHXII12:ATOMS

356430 The effective current due to the rotation of electron in atom depends on $Z$ and $n$ as

1

\frac{Z}{n^{2}}

2

\frac{Z}{n}

3

\frac{Z^{2}}{n^{3}}

4

\frac{Z^{2}}{n}

Explanation:

$i = \frac{e}{T} = \frac{e v_{n}}{2 π r_{n}} = \frac{m Z^{2} e^{5}}{4 ε_{0}^{2} h^{3} n^{3}}$

PHXII12:ATOMS

356431 The ratio of magnetic dipole moment to angular momentum in a hydrogen atom is

1

\frac{e}{m}

2

\frac{e}{2 m}

3

\frac{e}{3 m}

4

\frac{2 e}{m}

Explanation:

Magnetic moment
$μ = i A = \frac{q}{T} π r^{2} = \frac{q ω r^{2}}{2}$
Angular momentum
$L = m r^{2} ω$
$\frac{μ}{L} = \frac{q}{2 m} = \frac{e}{2 m}$

PHXII12:ATOMS

356432 Let the $P E$ of hydrogen atom in the ground state be zero. Then its total energy in the first excited state will be

1

23.8 e V

2

27.2 e V

3

10.2 e V

4

12.6 e V

Explanation:

When the $P$ . $E$ is taken as zero (Old-Frame) at infinity then the values of $P$ . $E$ , $K$ . $E$ & $T$ . $E$ for $H$ -atom are shown as follows
$P . E = 2 T . E & K . E = - T . E$
supporting img
$T . E = \frac{- 13.6 e V}{n^{2}}$

The above relations between $P$ . $E$ , $T$ . $E$ & $K$ . $E$ are not valid in the new frame.
Note: If we change the reference point for $P$ . $E$ then the absolute $K$ . $E$ and difference in $T$ . $E$ must remain same.

PHXII12:ATOMS

356433 Statement A :
The total energy of an electron revolving in any stationary orbit is negative.
Statement B :
Kinetic energy of revolving electron can be $(- v e)$ also.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both Statements are correct.

4 Both Statements are incorrect.

Explanation:

Energy can have positive or negative values. In the atom the electrons are bound. The negative energy of the revolving electrons does not allow the electrons to leave the atom. Kinetic energy of the electron is always $(+) v e$ . So option (1) is correct.

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation: