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CHXI02:STRUCTURE OF ATOM

307158 The energy of electron in first Bohr’s orbit of H-atom is –13.6 eV. What will be its potential energy in $4^{t h}$ orbit ?

1

- 13 .6 e V

2

- 3 .4 e V

3

- 0 .85 e V

4

- 1 .70 e V

Explanation:

$E_{n}$ for $H = \frac{- 13 .6 e V}{n^{2}}$
$E_{4}$ for $H = \frac{- 13 .6 e V}{4^{2}} = - 0 .85 e V$

CHXI02:STRUCTURE OF ATOM

307160 If m and e are the mass and charge of the revolving electron in the orbit of radius r for hydrogen atom, the total energy of the revolving electron will be:

1

\frac{1}{2} \frac{e^{2}}{r}

2

- \frac{e^{2}}{r}

3

\frac{m e^{2}}{r}

4

- \frac{1}{2} \frac{e^{2}}{r}

Explanation:

Total energy of a revolving electron is the sum of its kinetic and potential energy.
Total energy $= K . E + P . E .$
$= \frac{e^{2}}{2 r} + (- \frac{e^{2}}{r}) = - \frac{e^{2}}{2 r}$

JEE - 2014

CHXI02:STRUCTURE OF ATOM

307131 According to Bohr’s atomic theory, the correct statement is

1 Potential energy of electron

α \frac{Z^{2}}{n^{2}}

2 The product of velocity of electron and principal quantum number (n)

α Z^{2}

3 Frequency of revolution of electron in an orbit

α \frac{Z^{2}}{n^{3}}

4 Coulombic force of attraction on the electron

α \frac{Z^{2}}{n^{2}}

Explanation:

$v \propto \frac{Z}{n}; r \propto \frac{n^{2}}{Z}$
$(1) P . E . \propto \frac{1}{r} \propto \frac{Z}{n^{2}}$
$(2) v n \propto Z$
(3) Frequency of revolution
$= \frac{v}{2 π r_{n}} = \frac{Z / n}{2 π (n^{2} / Z)} = \frac{Z^{2}}{2 π n^{3}}$
Frequency $\propto \frac{n^{2}}{Z^{3}}$
(4) Coulombic force of attraction
$= \frac{Z e^{2}}{(4 π ε_{0}) r^{2}} \propto \frac{Z}{r^{2}} = \frac{Z}{n^{4} / Z} = \frac{Z^{2}}{n^{4}}$

CHXI02:STRUCTURE OF ATOM

307132 The amount of energy required to remove the electron from a $L i^{2 +}$ ion in its ground state is how many times greater than the amount of energy needed to remove the electron from an H atom in its ground state?

1

2

2

9

3

4

4

6

Explanation:

For $L i^{2 +}$ ion and H atom in their ground state $n = 1$ .
$\frac{{(E_{1})}_{L i^{2 +}}}{{(E_{1})}_{H}} = \frac{- \frac{1312 \times (3)^{2}}{{(1)}^{2}} k J / m o l}{\frac{1312 \times (1)^{2}}{{(1)}^{2}} k J / m o l} = 9$

CHXI02:STRUCTURE OF ATOM

307133 The ionisation energy of $H e^{+}$ is $19 .6 \times 1 0^{- 18} J a t o m^{- 1}$ . The energy of the first stationary state of $L i^{+ 2}$ will be

1

21 .2 \times 1 0^{- 18} J / a t o m

2

44 .10 \times 1 0^{- 18} J / a t o m

3

63 .2 \times 1 0^{- 18} J / a t o m

4

84 .2 \times 1 0^{- 18} J / a t o m

Explanation:

$E_{1}$ for $L i^{+ 2} = E_{1}$ for $H \times Z^{2} = E_{1}$ for $H \times 9$
$E_{1}$ for $H e^{+} = E_{1}$ for $H \times {Z^{2}}_{H e} = E_{1}$ for $H \times 4$
$E_{1}$ for $L i^{+ 2} = \frac{9}{4} E_{1}$ for $H e^{+}$
$= 19 .6 \times 1 0^{- 18} \times \frac{9}{4}$
$= 44 .10 \times 1 0^{- 18} J / a t o m$

CHXI02:STRUCTURE OF ATOM

307158 The energy of electron in first Bohr’s orbit of H-atom is –13.6 eV. What will be its potential energy in $4^{t h}$ orbit ?

1

- 13 .6 e V

2

- 3 .4 e V

3

- 0 .85 e V

4

- 1 .70 e V

Explanation:

$E_{n}$ for $H = \frac{- 13 .6 e V}{n^{2}}$
$E_{4}$ for $H = \frac{- 13 .6 e V}{4^{2}} = - 0 .85 e V$

CHXI02:STRUCTURE OF ATOM

307160 If m and e are the mass and charge of the revolving electron in the orbit of radius r for hydrogen atom, the total energy of the revolving electron will be:

1

\frac{1}{2} \frac{e^{2}}{r}

2

- \frac{e^{2}}{r}

3

\frac{m e^{2}}{r}

4

- \frac{1}{2} \frac{e^{2}}{r}

Explanation:

Total energy of a revolving electron is the sum of its kinetic and potential energy.
Total energy $= K . E + P . E .$
$= \frac{e^{2}}{2 r} + (- \frac{e^{2}}{r}) = - \frac{e^{2}}{2 r}$

JEE - 2014

CHXI02:STRUCTURE OF ATOM

307131 According to Bohr’s atomic theory, the correct statement is

1 Potential energy of electron

α \frac{Z^{2}}{n^{2}}

2 The product of velocity of electron and principal quantum number (n)

α Z^{2}

3 Frequency of revolution of electron in an orbit

α \frac{Z^{2}}{n^{3}}

4 Coulombic force of attraction on the electron

α \frac{Z^{2}}{n^{2}}

Explanation:

$v \propto \frac{Z}{n}; r \propto \frac{n^{2}}{Z}$
$(1) P . E . \propto \frac{1}{r} \propto \frac{Z}{n^{2}}$
$(2) v n \propto Z$
(3) Frequency of revolution
$= \frac{v}{2 π r_{n}} = \frac{Z / n}{2 π (n^{2} / Z)} = \frac{Z^{2}}{2 π n^{3}}$
Frequency $\propto \frac{n^{2}}{Z^{3}}$
(4) Coulombic force of attraction
$= \frac{Z e^{2}}{(4 π ε_{0}) r^{2}} \propto \frac{Z}{r^{2}} = \frac{Z}{n^{4} / Z} = \frac{Z^{2}}{n^{4}}$

CHXI02:STRUCTURE OF ATOM

307132 The amount of energy required to remove the electron from a $L i^{2 +}$ ion in its ground state is how many times greater than the amount of energy needed to remove the electron from an H atom in its ground state?

1

2

2

9

3

4

4

6

Explanation:

For $L i^{2 +}$ ion and H atom in their ground state $n = 1$ .
$\frac{{(E_{1})}_{L i^{2 +}}}{{(E_{1})}_{H}} = \frac{- \frac{1312 \times (3)^{2}}{{(1)}^{2}} k J / m o l}{\frac{1312 \times (1)^{2}}{{(1)}^{2}} k J / m o l} = 9$

CHXI02:STRUCTURE OF ATOM

307133 The ionisation energy of $H e^{+}$ is $19 .6 \times 1 0^{- 18} J a t o m^{- 1}$ . The energy of the first stationary state of $L i^{+ 2}$ will be

1

21 .2 \times 1 0^{- 18} J / a t o m

2

44 .10 \times 1 0^{- 18} J / a t o m

3

63 .2 \times 1 0^{- 18} J / a t o m

4

84 .2 \times 1 0^{- 18} J / a t o m

Explanation:

$E_{1}$ for $L i^{+ 2} = E_{1}$ for $H \times Z^{2} = E_{1}$ for $H \times 9$
$E_{1}$ for $H e^{+} = E_{1}$ for $H \times {Z^{2}}_{H e} = E_{1}$ for $H \times 4$
$E_{1}$ for $L i^{+ 2} = \frac{9}{4} E_{1}$ for $H e^{+}$
$= 19 .6 \times 1 0^{- 18} \times \frac{9}{4}$
$= 44 .10 \times 1 0^{- 18} J / a t o m$

CHXI02:STRUCTURE OF ATOM

307158 The energy of electron in first Bohr’s orbit of H-atom is –13.6 eV. What will be its potential energy in $4^{t h}$ orbit ?

1

- 13 .6 e V

2

- 3 .4 e V

3

- 0 .85 e V

4

- 1 .70 e V

Explanation:

$E_{n}$ for $H = \frac{- 13 .6 e V}{n^{2}}$
$E_{4}$ for $H = \frac{- 13 .6 e V}{4^{2}} = - 0 .85 e V$

CHXI02:STRUCTURE OF ATOM

307160 If m and e are the mass and charge of the revolving electron in the orbit of radius r for hydrogen atom, the total energy of the revolving electron will be:

1

\frac{1}{2} \frac{e^{2}}{r}

2

- \frac{e^{2}}{r}

3

\frac{m e^{2}}{r}

4

- \frac{1}{2} \frac{e^{2}}{r}

Explanation:

Total energy of a revolving electron is the sum of its kinetic and potential energy.
Total energy $= K . E + P . E .$
$= \frac{e^{2}}{2 r} + (- \frac{e^{2}}{r}) = - \frac{e^{2}}{2 r}$

JEE - 2014

CHXI02:STRUCTURE OF ATOM

307131 According to Bohr’s atomic theory, the correct statement is

1 Potential energy of electron

α \frac{Z^{2}}{n^{2}}

2 The product of velocity of electron and principal quantum number (n)

α Z^{2}

3 Frequency of revolution of electron in an orbit

α \frac{Z^{2}}{n^{3}}

4 Coulombic force of attraction on the electron

α \frac{Z^{2}}{n^{2}}

Explanation:

$v \propto \frac{Z}{n}; r \propto \frac{n^{2}}{Z}$
$(1) P . E . \propto \frac{1}{r} \propto \frac{Z}{n^{2}}$
$(2) v n \propto Z$
(3) Frequency of revolution
$= \frac{v}{2 π r_{n}} = \frac{Z / n}{2 π (n^{2} / Z)} = \frac{Z^{2}}{2 π n^{3}}$
Frequency $\propto \frac{n^{2}}{Z^{3}}$
(4) Coulombic force of attraction
$= \frac{Z e^{2}}{(4 π ε_{0}) r^{2}} \propto \frac{Z}{r^{2}} = \frac{Z}{n^{4} / Z} = \frac{Z^{2}}{n^{4}}$

CHXI02:STRUCTURE OF ATOM

307132 The amount of energy required to remove the electron from a $L i^{2 +}$ ion in its ground state is how many times greater than the amount of energy needed to remove the electron from an H atom in its ground state?

1

2

2

9

3

4

4

6

Explanation:

For $L i^{2 +}$ ion and H atom in their ground state $n = 1$ .
$\frac{{(E_{1})}_{L i^{2 +}}}{{(E_{1})}_{H}} = \frac{- \frac{1312 \times (3)^{2}}{{(1)}^{2}} k J / m o l}{\frac{1312 \times (1)^{2}}{{(1)}^{2}} k J / m o l} = 9$

CHXI02:STRUCTURE OF ATOM

307133 The ionisation energy of $H e^{+}$ is $19 .6 \times 1 0^{- 18} J a t o m^{- 1}$ . The energy of the first stationary state of $L i^{+ 2}$ will be

1

21 .2 \times 1 0^{- 18} J / a t o m

2

44 .10 \times 1 0^{- 18} J / a t o m

3

63 .2 \times 1 0^{- 18} J / a t o m

4

84 .2 \times 1 0^{- 18} J / a t o m

Explanation:

$E_{1}$ for $L i^{+ 2} = E_{1}$ for $H \times Z^{2} = E_{1}$ for $H \times 9$
$E_{1}$ for $H e^{+} = E_{1}$ for $H \times {Z^{2}}_{H e} = E_{1}$ for $H \times 4$
$E_{1}$ for $L i^{+ 2} = \frac{9}{4} E_{1}$ for $H e^{+}$
$= 19 .6 \times 1 0^{- 18} \times \frac{9}{4}$
$= 44 .10 \times 1 0^{- 18} J / a t o m$

CHXI02:STRUCTURE OF ATOM

307158 The energy of electron in first Bohr’s orbit of H-atom is –13.6 eV. What will be its potential energy in $4^{t h}$ orbit ?

1

- 13 .6 e V

2

- 3 .4 e V

3

- 0 .85 e V

4

- 1 .70 e V

Explanation:

$E_{n}$ for $H = \frac{- 13 .6 e V}{n^{2}}$
$E_{4}$ for $H = \frac{- 13 .6 e V}{4^{2}} = - 0 .85 e V$

CHXI02:STRUCTURE OF ATOM

307160 If m and e are the mass and charge of the revolving electron in the orbit of radius r for hydrogen atom, the total energy of the revolving electron will be:

1

\frac{1}{2} \frac{e^{2}}{r}

2

- \frac{e^{2}}{r}

3

\frac{m e^{2}}{r}

4

- \frac{1}{2} \frac{e^{2}}{r}

Explanation:

Total energy of a revolving electron is the sum of its kinetic and potential energy.
Total energy $= K . E + P . E .$
$= \frac{e^{2}}{2 r} + (- \frac{e^{2}}{r}) = - \frac{e^{2}}{2 r}$

JEE - 2014

CHXI02:STRUCTURE OF ATOM

307131 According to Bohr’s atomic theory, the correct statement is

1 Potential energy of electron

α \frac{Z^{2}}{n^{2}}

2 The product of velocity of electron and principal quantum number (n)

α Z^{2}

3 Frequency of revolution of electron in an orbit

α \frac{Z^{2}}{n^{3}}

4 Coulombic force of attraction on the electron

α \frac{Z^{2}}{n^{2}}

Explanation:

$v \propto \frac{Z}{n}; r \propto \frac{n^{2}}{Z}$
$(1) P . E . \propto \frac{1}{r} \propto \frac{Z}{n^{2}}$
$(2) v n \propto Z$
(3) Frequency of revolution
$= \frac{v}{2 π r_{n}} = \frac{Z / n}{2 π (n^{2} / Z)} = \frac{Z^{2}}{2 π n^{3}}$
Frequency $\propto \frac{n^{2}}{Z^{3}}$
(4) Coulombic force of attraction
$= \frac{Z e^{2}}{(4 π ε_{0}) r^{2}} \propto \frac{Z}{r^{2}} = \frac{Z}{n^{4} / Z} = \frac{Z^{2}}{n^{4}}$

CHXI02:STRUCTURE OF ATOM

307132 The amount of energy required to remove the electron from a $L i^{2 +}$ ion in its ground state is how many times greater than the amount of energy needed to remove the electron from an H atom in its ground state?

1

2

2

9

3

4

4

6

Explanation:

For $L i^{2 +}$ ion and H atom in their ground state $n = 1$ .
$\frac{{(E_{1})}_{L i^{2 +}}}{{(E_{1})}_{H}} = \frac{- \frac{1312 \times (3)^{2}}{{(1)}^{2}} k J / m o l}{\frac{1312 \times (1)^{2}}{{(1)}^{2}} k J / m o l} = 9$

CHXI02:STRUCTURE OF ATOM

307133 The ionisation energy of $H e^{+}$ is $19 .6 \times 1 0^{- 18} J a t o m^{- 1}$ . The energy of the first stationary state of $L i^{+ 2}$ will be

1

21 .2 \times 1 0^{- 18} J / a t o m

2

44 .10 \times 1 0^{- 18} J / a t o m

3

63 .2 \times 1 0^{- 18} J / a t o m

4

84 .2 \times 1 0^{- 18} J / a t o m

Explanation:

$E_{1}$ for $L i^{+ 2} = E_{1}$ for $H \times Z^{2} = E_{1}$ for $H \times 9$
$E_{1}$ for $H e^{+} = E_{1}$ for $H \times {Z^{2}}_{H e} = E_{1}$ for $H \times 4$
$E_{1}$ for $L i^{+ 2} = \frac{9}{4} E_{1}$ for $H e^{+}$
$= 19 .6 \times 1 0^{- 18} \times \frac{9}{4}$
$= 44 .10 \times 1 0^{- 18} J / a t o m$

CHXI02:STRUCTURE OF ATOM

307158 The energy of electron in first Bohr’s orbit of H-atom is –13.6 eV. What will be its potential energy in $4^{t h}$ orbit ?

1

- 13 .6 e V

2

- 3 .4 e V

3

- 0 .85 e V

4

- 1 .70 e V

Explanation:

$E_{n}$ for $H = \frac{- 13 .6 e V}{n^{2}}$
$E_{4}$ for $H = \frac{- 13 .6 e V}{4^{2}} = - 0 .85 e V$

CHXI02:STRUCTURE OF ATOM

307160 If m and e are the mass and charge of the revolving electron in the orbit of radius r for hydrogen atom, the total energy of the revolving electron will be:

1

\frac{1}{2} \frac{e^{2}}{r}

2

- \frac{e^{2}}{r}

3

\frac{m e^{2}}{r}

4

- \frac{1}{2} \frac{e^{2}}{r}

Explanation:

Total energy of a revolving electron is the sum of its kinetic and potential energy.
Total energy $= K . E + P . E .$
$= \frac{e^{2}}{2 r} + (- \frac{e^{2}}{r}) = - \frac{e^{2}}{2 r}$

JEE - 2014

CHXI02:STRUCTURE OF ATOM

307131 According to Bohr’s atomic theory, the correct statement is

1 Potential energy of electron

α \frac{Z^{2}}{n^{2}}

2 The product of velocity of electron and principal quantum number (n)

α Z^{2}

3 Frequency of revolution of electron in an orbit

α \frac{Z^{2}}{n^{3}}

4 Coulombic force of attraction on the electron

α \frac{Z^{2}}{n^{2}}

Explanation:

$v \propto \frac{Z}{n}; r \propto \frac{n^{2}}{Z}$
$(1) P . E . \propto \frac{1}{r} \propto \frac{Z}{n^{2}}$
$(2) v n \propto Z$
(3) Frequency of revolution
$= \frac{v}{2 π r_{n}} = \frac{Z / n}{2 π (n^{2} / Z)} = \frac{Z^{2}}{2 π n^{3}}$
Frequency $\propto \frac{n^{2}}{Z^{3}}$
(4) Coulombic force of attraction
$= \frac{Z e^{2}}{(4 π ε_{0}) r^{2}} \propto \frac{Z}{r^{2}} = \frac{Z}{n^{4} / Z} = \frac{Z^{2}}{n^{4}}$

CHXI02:STRUCTURE OF ATOM

307132 The amount of energy required to remove the electron from a $L i^{2 +}$ ion in its ground state is how many times greater than the amount of energy needed to remove the electron from an H atom in its ground state?

1

2

2

9

3

4

4

6

Explanation:

For $L i^{2 +}$ ion and H atom in their ground state $n = 1$ .
$\frac{{(E_{1})}_{L i^{2 +}}}{{(E_{1})}_{H}} = \frac{- \frac{1312 \times (3)^{2}}{{(1)}^{2}} k J / m o l}{\frac{1312 \times (1)^{2}}{{(1)}^{2}} k J / m o l} = 9$

CHXI02:STRUCTURE OF ATOM

307133 The ionisation energy of $H e^{+}$ is $19 .6 \times 1 0^{- 18} J a t o m^{- 1}$ . The energy of the first stationary state of $L i^{+ 2}$ will be

1

21 .2 \times 1 0^{- 18} J / a t o m

2

44 .10 \times 1 0^{- 18} J / a t o m

3

63 .2 \times 1 0^{- 18} J / a t o m

4

84 .2 \times 1 0^{- 18} J / a t o m

Explanation:

$E_{1}$ for $L i^{+ 2} = E_{1}$ for $H \times Z^{2} = E_{1}$ for $H \times 9$
$E_{1}$ for $H e^{+} = E_{1}$ for $H \times {Z^{2}}_{H e} = E_{1}$ for $H \times 4$
$E_{1}$ for $L i^{+ 2} = \frac{9}{4} E_{1}$ for $H e^{+}$
$= 19 .6 \times 1 0^{- 18} \times \frac{9}{4}$
$= 44 .10 \times 1 0^{- 18} J / a t o m$