QBManager

PHXII12:ATOMS

356438 In hydrogen atom, electron excites from ground state to higher energy state and its orbital velocity is reduced to ${(\frac{1}{3})}^{r d}$ of its initial value. The radius of the orbit in the ground state is $R$ . The radius of the orbit in that higher state is ......

1

2 R

2

3 R

3

27 R

4

9 R

Explanation:

As per problem,
$\frac{v_{h}}{v_{g}} = \frac{1}{3} (1)$
Where subscripts $h$ and $g$ denotes higher energy state and ground state.
Orbital velocity of electron in the $n^{t h}$ orbit is
$v_{n} = \frac{e^{2}}{2 ε_{0} n h}$ or $v_{n} \propto \frac{1}{n}$
For ground state, $n = 1$
$∴ \frac{v_{h}}{v_{g}} = \frac{1}{n} (2)$
Equating (1) and (2), we get $n = 3$ (higher energy state)
Radius of $n^{t h}$ orbit is
$r_{n} = \frac{n^{2} h^{2} ε_{0}}{π e^{2} m}$ or $r_{n} \propto n^{2}$
$∴ \frac{r_{3}}{r_{1}} = \frac{{(3)}^{2}}{{(1)}^{2}} = 9$
$r_{3} = 9 r_{1} = 9 R$ $(∵ given r_{1} = R)$

PHXII12:ATOMS

356439 Assertion :
The electron in the hydrogen atom passes from energy level $n = 4$ to the $n = 1$ level. The maximum and minimum number of photons that can be emitted are six and one respectively.
Reason :
The photons are emitted when electron make a transition from the lower energy state to the higher energy state.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

The maximum number of photon emissions
$N_{max} = \frac{n (n - 1)}{2}$
$= \frac{4 (4 - 1)}{2} = 6$
Photons are emitted when electron have a transition from higher to lower state. The minimum is 1 for a direct jump. So correct option is (3).

PHXII12:ATOMS

356440 The average $K E$ of molecules in a gas at temperature $T$ is $\frac{3}{2} K T$ .Find the temperature at which the average $K E$ of molecules equal to binding energy of $H$ -atom

1

1 .05 \times 1 0^{4} K

2

1 .05 \times 1 0^{5} K

3

1 .05 \times 1 0^{3} K

4

N o n e o f t h e s e

Explanation:

$\frac{3}{2} K T = 13.6 e V$
$\Rightarrow \frac{3}{2} \times \frac{1.38 \times 10^{- 23}}{1.6 \times 10^{- 19}} \times T = 13.6$
$T = 1.05 \times 10^{5} K$

PHXII12:ATOMS

356441 For an electron in the seventh orbit of hydrogen, the moment of momentum as per Bohr’s model is

1

\frac{7 h}{2 π}

2

\frac{2 h}{π}

3

\frac{5 h}{2 π}

4

2 π h

Explanation:

We know from Bohr’s theory that in an orbit angular momentum $= n \cdot \frac{h}{2 π}$
For second orbit $n = 7$
So, angular momentum $= 7 \times \frac{h}{2 π} = \frac{7 h}{2 π}$

PHXII12:ATOMS

356438 In hydrogen atom, electron excites from ground state to higher energy state and its orbital velocity is reduced to ${(\frac{1}{3})}^{r d}$ of its initial value. The radius of the orbit in the ground state is $R$ . The radius of the orbit in that higher state is ......

1

2 R

2

3 R

3

27 R

4

9 R

Explanation:

As per problem,
$\frac{v_{h}}{v_{g}} = \frac{1}{3} (1)$
Where subscripts $h$ and $g$ denotes higher energy state and ground state.
Orbital velocity of electron in the $n^{t h}$ orbit is
$v_{n} = \frac{e^{2}}{2 ε_{0} n h}$ or $v_{n} \propto \frac{1}{n}$
For ground state, $n = 1$
$∴ \frac{v_{h}}{v_{g}} = \frac{1}{n} (2)$
Equating (1) and (2), we get $n = 3$ (higher energy state)
Radius of $n^{t h}$ orbit is
$r_{n} = \frac{n^{2} h^{2} ε_{0}}{π e^{2} m}$ or $r_{n} \propto n^{2}$
$∴ \frac{r_{3}}{r_{1}} = \frac{{(3)}^{2}}{{(1)}^{2}} = 9$
$r_{3} = 9 r_{1} = 9 R$ $(∵ given r_{1} = R)$

PHXII12:ATOMS

356439 Assertion :
The electron in the hydrogen atom passes from energy level $n = 4$ to the $n = 1$ level. The maximum and minimum number of photons that can be emitted are six and one respectively.
Reason :
The photons are emitted when electron make a transition from the lower energy state to the higher energy state.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

The maximum number of photon emissions
$N_{max} = \frac{n (n - 1)}{2}$
$= \frac{4 (4 - 1)}{2} = 6$
Photons are emitted when electron have a transition from higher to lower state. The minimum is 1 for a direct jump. So correct option is (3).

PHXII12:ATOMS

356440 The average $K E$ of molecules in a gas at temperature $T$ is $\frac{3}{2} K T$ .Find the temperature at which the average $K E$ of molecules equal to binding energy of $H$ -atom

1

1 .05 \times 1 0^{4} K

2

1 .05 \times 1 0^{5} K

3

1 .05 \times 1 0^{3} K

4

N o n e o f t h e s e

Explanation:

$\frac{3}{2} K T = 13.6 e V$
$\Rightarrow \frac{3}{2} \times \frac{1.38 \times 10^{- 23}}{1.6 \times 10^{- 19}} \times T = 13.6$
$T = 1.05 \times 10^{5} K$

PHXII12:ATOMS

356441 For an electron in the seventh orbit of hydrogen, the moment of momentum as per Bohr’s model is

1

\frac{7 h}{2 π}

2

\frac{2 h}{π}

3

\frac{5 h}{2 π}

4

2 π h

Explanation:

We know from Bohr’s theory that in an orbit angular momentum $= n \cdot \frac{h}{2 π}$
For second orbit $n = 7$
So, angular momentum $= 7 \times \frac{h}{2 π} = \frac{7 h}{2 π}$

PHXII12:ATOMS

356438 In hydrogen atom, electron excites from ground state to higher energy state and its orbital velocity is reduced to ${(\frac{1}{3})}^{r d}$ of its initial value. The radius of the orbit in the ground state is $R$ . The radius of the orbit in that higher state is ......

1

2 R

2

3 R

3

27 R

4

9 R

Explanation:

As per problem,
$\frac{v_{h}}{v_{g}} = \frac{1}{3} (1)$
Where subscripts $h$ and $g$ denotes higher energy state and ground state.
Orbital velocity of electron in the $n^{t h}$ orbit is
$v_{n} = \frac{e^{2}}{2 ε_{0} n h}$ or $v_{n} \propto \frac{1}{n}$
For ground state, $n = 1$
$∴ \frac{v_{h}}{v_{g}} = \frac{1}{n} (2)$
Equating (1) and (2), we get $n = 3$ (higher energy state)
Radius of $n^{t h}$ orbit is
$r_{n} = \frac{n^{2} h^{2} ε_{0}}{π e^{2} m}$ or $r_{n} \propto n^{2}$
$∴ \frac{r_{3}}{r_{1}} = \frac{{(3)}^{2}}{{(1)}^{2}} = 9$
$r_{3} = 9 r_{1} = 9 R$ $(∵ given r_{1} = R)$

PHXII12:ATOMS

356439 Assertion :
The electron in the hydrogen atom passes from energy level $n = 4$ to the $n = 1$ level. The maximum and minimum number of photons that can be emitted are six and one respectively.
Reason :
The photons are emitted when electron make a transition from the lower energy state to the higher energy state.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

The maximum number of photon emissions
$N_{max} = \frac{n (n - 1)}{2}$
$= \frac{4 (4 - 1)}{2} = 6$
Photons are emitted when electron have a transition from higher to lower state. The minimum is 1 for a direct jump. So correct option is (3).

PHXII12:ATOMS

356440 The average $K E$ of molecules in a gas at temperature $T$ is $\frac{3}{2} K T$ .Find the temperature at which the average $K E$ of molecules equal to binding energy of $H$ -atom

1

1 .05 \times 1 0^{4} K

2

1 .05 \times 1 0^{5} K

3

1 .05 \times 1 0^{3} K

4

N o n e o f t h e s e

Explanation:

$\frac{3}{2} K T = 13.6 e V$
$\Rightarrow \frac{3}{2} \times \frac{1.38 \times 10^{- 23}}{1.6 \times 10^{- 19}} \times T = 13.6$
$T = 1.05 \times 10^{5} K$

PHXII12:ATOMS

356441 For an electron in the seventh orbit of hydrogen, the moment of momentum as per Bohr’s model is

1

\frac{7 h}{2 π}

2

\frac{2 h}{π}

3

\frac{5 h}{2 π}

4

2 π h

Explanation:

We know from Bohr’s theory that in an orbit angular momentum $= n \cdot \frac{h}{2 π}$
For second orbit $n = 7$
So, angular momentum $= 7 \times \frac{h}{2 π} = \frac{7 h}{2 π}$

PHXII12:ATOMS

356438 In hydrogen atom, electron excites from ground state to higher energy state and its orbital velocity is reduced to ${(\frac{1}{3})}^{r d}$ of its initial value. The radius of the orbit in the ground state is $R$ . The radius of the orbit in that higher state is ......

1

2 R

2

3 R

3

27 R

4

9 R

Explanation:

As per problem,
$\frac{v_{h}}{v_{g}} = \frac{1}{3} (1)$
Where subscripts $h$ and $g$ denotes higher energy state and ground state.
Orbital velocity of electron in the $n^{t h}$ orbit is
$v_{n} = \frac{e^{2}}{2 ε_{0} n h}$ or $v_{n} \propto \frac{1}{n}$
For ground state, $n = 1$
$∴ \frac{v_{h}}{v_{g}} = \frac{1}{n} (2)$
Equating (1) and (2), we get $n = 3$ (higher energy state)
Radius of $n^{t h}$ orbit is
$r_{n} = \frac{n^{2} h^{2} ε_{0}}{π e^{2} m}$ or $r_{n} \propto n^{2}$
$∴ \frac{r_{3}}{r_{1}} = \frac{{(3)}^{2}}{{(1)}^{2}} = 9$
$r_{3} = 9 r_{1} = 9 R$ $(∵ given r_{1} = R)$

PHXII12:ATOMS

356439 Assertion :
The electron in the hydrogen atom passes from energy level $n = 4$ to the $n = 1$ level. The maximum and minimum number of photons that can be emitted are six and one respectively.
Reason :
The photons are emitted when electron make a transition from the lower energy state to the higher energy state.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

The maximum number of photon emissions
$N_{max} = \frac{n (n - 1)}{2}$
$= \frac{4 (4 - 1)}{2} = 6$
Photons are emitted when electron have a transition from higher to lower state. The minimum is 1 for a direct jump. So correct option is (3).

PHXII12:ATOMS

356440 The average $K E$ of molecules in a gas at temperature $T$ is $\frac{3}{2} K T$ .Find the temperature at which the average $K E$ of molecules equal to binding energy of $H$ -atom

1

1 .05 \times 1 0^{4} K

2

1 .05 \times 1 0^{5} K

3

1 .05 \times 1 0^{3} K

4

N o n e o f t h e s e

Explanation:

$\frac{3}{2} K T = 13.6 e V$
$\Rightarrow \frac{3}{2} \times \frac{1.38 \times 10^{- 23}}{1.6 \times 10^{- 19}} \times T = 13.6$
$T = 1.05 \times 10^{5} K$

PHXII12:ATOMS

356441 For an electron in the seventh orbit of hydrogen, the moment of momentum as per Bohr’s model is

1

\frac{7 h}{2 π}

2

\frac{2 h}{π}

3

\frac{5 h}{2 π}

4

2 π h

Explanation:

We know from Bohr’s theory that in an orbit angular momentum $= n \cdot \frac{h}{2 π}$
For second orbit $n = 7$
So, angular momentum $= 7 \times \frac{h}{2 π} = \frac{7 h}{2 π}$

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation: