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PHXII12:ATOMS

356376 An electron with $K E 6 e V$ is incident on a hydrogen atom in its ground state. The collision

1 May be partially inelastic

2 Must be completely inelastic

3 May be partially elastic

4 Must be elastic

Explanation:

The minimum energy required for excitation of hydrogen atom is 10.2 $e V$ . Here the $K$ . $E$ of the incoming electron is less than 10.2 $e V$ . So excitation does not take place. The $K$ . $E$ of the electron will be retained as $K$ . $E$ for the system. The collision must be perfectly elastic.

PHXII12:ATOMS

356377 A moving hydrogen atom makes a head-on collision with a stationary hydrogen atom. Before collision, both atoms are in ground state and after collision they move together. What is the minimum value of the kinetic energy of the moving hydrogen atom, such that one of the atoms reaches one of the excitation state?
supporting img

1

20.4 e V

2

10.2 e V

3

30.6 e V

4

A n y v a l u e

Explanation:

Method-1
From conservation of linear momentum
$m u + 0 = (m + m) v \Rightarrow v = \frac{u}{2};$
In this case loss in $K E$ will be maximum. Now this $Δ K_{max}$ must be greater than the energy required to excite
$| {(Δ K)}_{max} | = | \frac{1}{2} 2 m {(\frac{u}{2})}^{2} - \frac{1}{2} m u^{2} | = \frac{1}{4} m u^{2}$
$\frac{1}{4} m u^{2} \geq 10.2 e v \Rightarrow \frac{1}{2} m u^{2} \geq 20.4 e V$
Therefore, the minimum kinetic energy of moving hydrogen is $20.4 e V$ .
Method-2
The threshold energy ( $K . E_{min}$ ) required for the moving $H$ -atom is
$K . E_{T h} = E_{a} [1 + \frac{m}{M}]$
$E_{a} -$ Energy of excitation, $m$ - mass of moving particle
$M -$ Mass of particle at rest
$\Rightarrow K . E_{T h} = 10.2 e V [1 + \frac{1}{1}] = 20.4 e V$

PHXII12:ATOMS

356378 A hydrogen atom moves with a velocity $u$ and makes a head on inelastic collision are in the ground state before collision. What is the minimum value of $u$ , if one of them is to be given a minimum excitation energy? The ionization energy is $13.6 e V$ . Mass of the hydrogen atom is $1.0078 \times 1.66 \times 10^{- 27} k g$ .

1

3.74 \times 10^{4} m / s

2

7.35 \times 10^{4} m / s

3

1.64 \times 10^{4} m / s

4

6.24 \times 10^{4} m / s

Explanation:

From conservation of momentum $m u = 2 m v$
$\Rightarrow v = u / 2 (1)$
The energy of excitation $Δ E$ is given by
$Δ E = \frac{1}{2} m u^{2} - 2 \times \frac{1}{2} m (u / 2)^{2}$
$\Rightarrow Δ E = \frac{1}{4} m u^{2} (2)$
The minimum excitation energy for a $H$ -atom is for transition $n_{1} = 1$ to $n_{2} = 2$ which corresponds to an energy of $10.2 e V$ . From Eq. (2)
$\frac{1}{4} \times (1.0078 \times 1.66 \times 10^{- 27}) u^{2}$
$= 10.2 \times (1.6 \times 10^{- 19})$
$\Rightarrow u = 6.24 \times 10^{4} m / s$

PHXII12:ATOMS

356379 A $H e^{+}$ ion is at rest and is in ground state. A neutron with intial kinetic energy $K$ collides head on with the $H e^{+}$ ion. Find minimum value of $K$ so that there can be an inelastic collision between these two particles.
supporting img

1

51 e V

2

30.6 e V

3

20.4 e V

4

40.8 e V

Explanation:

Method-1
From conservation of linear momentum
$m u = (m + 4 m) v \Rightarrow v = \frac{u}{5};$
In this case loss in $K E$ will be maximum. Now this $Δ K_{max}$ must be greater than the energy received to excite
$| Δ K_{max} | = \frac{1}{2} m u^{2} - \frac{1}{2} (5 m) {(\frac{u}{5})}^{2} = \frac{1}{2} m u^{2} (1 - \frac{1}{5})$
$= \frac{1}{2} m u^{2} \cdot \frac{4}{5}$
$∴ \frac{1}{2} m u^{2} \cdot \frac{4}{5} \geq 40.8 \Rightarrow \frac{1}{2} m u^{2} \geq 51 e V$
Hence, minimum value of $K$ for an inelastic collision: $K_{min} = 51 e V .$
Method-2
The Thereshold energy
$K . E_{T h} = E_{a} [1 + \frac{m}{M}] (M \approx 4 m)$
$E_{1} = \frac{- 13.6 \times 2^{2}}{1^{2}}, E_{2} = \frac{- 13.6 \times 2^{2}}{2^{2}}$
$\Rightarrow E_{a} = E_{1} - E_{2}$
$\Rightarrow E_{a} = 13.6 \times 3 e V \Rightarrow K . E_{T h} = \frac{5 E_{a}}{4} = 51 e V$

PHXII12:ATOMS

356380 Consider a hydrogen atom present at rest in ground state. A fast moving neutron collides with the atom. Find the threshold energy of the neutron for excitations $n_{1} = 2 t o n_{2} = 3$ $(m_{N} ≃ m_{H})$

1

24.2 e V

2

12.4 e V

3

10.2 e V

4

20.4 e V

Explanation:

The threshold energy
$E_{T h} = E_{a} [1 + \frac{m}{M}]$
$m = m_{n}$ & $M = m_{H}$
$E_{a} = 10.2 e V & m_{n} \approx m_{H}$
In the following table threshold energies are given for different excitations of $H$ -atom
It can be observed from the above table that $E_{T h} = 24.2 e V$
supporting img

PHXII12:ATOMS

356376 An electron with $K E 6 e V$ is incident on a hydrogen atom in its ground state. The collision

1 May be partially inelastic

2 Must be completely inelastic

3 May be partially elastic

4 Must be elastic

Explanation:

The minimum energy required for excitation of hydrogen atom is 10.2 $e V$ . Here the $K$ . $E$ of the incoming electron is less than 10.2 $e V$ . So excitation does not take place. The $K$ . $E$ of the electron will be retained as $K$ . $E$ for the system. The collision must be perfectly elastic.

PHXII12:ATOMS

356377 A moving hydrogen atom makes a head-on collision with a stationary hydrogen atom. Before collision, both atoms are in ground state and after collision they move together. What is the minimum value of the kinetic energy of the moving hydrogen atom, such that one of the atoms reaches one of the excitation state?
supporting img

1

20.4 e V

2

10.2 e V

3

30.6 e V

4

A n y v a l u e

Explanation:

Method-1
From conservation of linear momentum
$m u + 0 = (m + m) v \Rightarrow v = \frac{u}{2};$
In this case loss in $K E$ will be maximum. Now this $Δ K_{max}$ must be greater than the energy required to excite
$| {(Δ K)}_{max} | = | \frac{1}{2} 2 m {(\frac{u}{2})}^{2} - \frac{1}{2} m u^{2} | = \frac{1}{4} m u^{2}$
$\frac{1}{4} m u^{2} \geq 10.2 e v \Rightarrow \frac{1}{2} m u^{2} \geq 20.4 e V$
Therefore, the minimum kinetic energy of moving hydrogen is $20.4 e V$ .
Method-2
The threshold energy ( $K . E_{min}$ ) required for the moving $H$ -atom is
$K . E_{T h} = E_{a} [1 + \frac{m}{M}]$
$E_{a} -$ Energy of excitation, $m$ - mass of moving particle
$M -$ Mass of particle at rest
$\Rightarrow K . E_{T h} = 10.2 e V [1 + \frac{1}{1}] = 20.4 e V$

PHXII12:ATOMS

356378 A hydrogen atom moves with a velocity $u$ and makes a head on inelastic collision are in the ground state before collision. What is the minimum value of $u$ , if one of them is to be given a minimum excitation energy? The ionization energy is $13.6 e V$ . Mass of the hydrogen atom is $1.0078 \times 1.66 \times 10^{- 27} k g$ .

1

3.74 \times 10^{4} m / s

2

7.35 \times 10^{4} m / s

3

1.64 \times 10^{4} m / s

4

6.24 \times 10^{4} m / s

Explanation:

From conservation of momentum $m u = 2 m v$
$\Rightarrow v = u / 2 (1)$
The energy of excitation $Δ E$ is given by
$Δ E = \frac{1}{2} m u^{2} - 2 \times \frac{1}{2} m (u / 2)^{2}$
$\Rightarrow Δ E = \frac{1}{4} m u^{2} (2)$
The minimum excitation energy for a $H$ -atom is for transition $n_{1} = 1$ to $n_{2} = 2$ which corresponds to an energy of $10.2 e V$ . From Eq. (2)
$\frac{1}{4} \times (1.0078 \times 1.66 \times 10^{- 27}) u^{2}$
$= 10.2 \times (1.6 \times 10^{- 19})$
$\Rightarrow u = 6.24 \times 10^{4} m / s$

PHXII12:ATOMS

356379 A $H e^{+}$ ion is at rest and is in ground state. A neutron with intial kinetic energy $K$ collides head on with the $H e^{+}$ ion. Find minimum value of $K$ so that there can be an inelastic collision between these two particles.
supporting img

1

51 e V

2

30.6 e V

3

20.4 e V

4

40.8 e V

Explanation:

Method-1
From conservation of linear momentum
$m u = (m + 4 m) v \Rightarrow v = \frac{u}{5};$
In this case loss in $K E$ will be maximum. Now this $Δ K_{max}$ must be greater than the energy received to excite
$| Δ K_{max} | = \frac{1}{2} m u^{2} - \frac{1}{2} (5 m) {(\frac{u}{5})}^{2} = \frac{1}{2} m u^{2} (1 - \frac{1}{5})$
$= \frac{1}{2} m u^{2} \cdot \frac{4}{5}$
$∴ \frac{1}{2} m u^{2} \cdot \frac{4}{5} \geq 40.8 \Rightarrow \frac{1}{2} m u^{2} \geq 51 e V$
Hence, minimum value of $K$ for an inelastic collision: $K_{min} = 51 e V .$
Method-2
The Thereshold energy
$K . E_{T h} = E_{a} [1 + \frac{m}{M}] (M \approx 4 m)$
$E_{1} = \frac{- 13.6 \times 2^{2}}{1^{2}}, E_{2} = \frac{- 13.6 \times 2^{2}}{2^{2}}$
$\Rightarrow E_{a} = E_{1} - E_{2}$
$\Rightarrow E_{a} = 13.6 \times 3 e V \Rightarrow K . E_{T h} = \frac{5 E_{a}}{4} = 51 e V$

PHXII12:ATOMS

356380 Consider a hydrogen atom present at rest in ground state. A fast moving neutron collides with the atom. Find the threshold energy of the neutron for excitations $n_{1} = 2 t o n_{2} = 3$ $(m_{N} ≃ m_{H})$

1

24.2 e V

2

12.4 e V

3

10.2 e V

4

20.4 e V

Explanation:

The threshold energy
$E_{T h} = E_{a} [1 + \frac{m}{M}]$
$m = m_{n}$ & $M = m_{H}$
$E_{a} = 10.2 e V & m_{n} \approx m_{H}$
In the following table threshold energies are given for different excitations of $H$ -atom
It can be observed from the above table that $E_{T h} = 24.2 e V$
supporting img

PHXII12:ATOMS

356376 An electron with $K E 6 e V$ is incident on a hydrogen atom in its ground state. The collision

1 May be partially inelastic

2 Must be completely inelastic

3 May be partially elastic

4 Must be elastic

Explanation:

The minimum energy required for excitation of hydrogen atom is 10.2 $e V$ . Here the $K$ . $E$ of the incoming electron is less than 10.2 $e V$ . So excitation does not take place. The $K$ . $E$ of the electron will be retained as $K$ . $E$ for the system. The collision must be perfectly elastic.

PHXII12:ATOMS

356377 A moving hydrogen atom makes a head-on collision with a stationary hydrogen atom. Before collision, both atoms are in ground state and after collision they move together. What is the minimum value of the kinetic energy of the moving hydrogen atom, such that one of the atoms reaches one of the excitation state?
supporting img

1

20.4 e V

2

10.2 e V

3

30.6 e V

4

A n y v a l u e

Explanation:

Method-1
From conservation of linear momentum
$m u + 0 = (m + m) v \Rightarrow v = \frac{u}{2};$
In this case loss in $K E$ will be maximum. Now this $Δ K_{max}$ must be greater than the energy required to excite
$| {(Δ K)}_{max} | = | \frac{1}{2} 2 m {(\frac{u}{2})}^{2} - \frac{1}{2} m u^{2} | = \frac{1}{4} m u^{2}$
$\frac{1}{4} m u^{2} \geq 10.2 e v \Rightarrow \frac{1}{2} m u^{2} \geq 20.4 e V$
Therefore, the minimum kinetic energy of moving hydrogen is $20.4 e V$ .
Method-2
The threshold energy ( $K . E_{min}$ ) required for the moving $H$ -atom is
$K . E_{T h} = E_{a} [1 + \frac{m}{M}]$
$E_{a} -$ Energy of excitation, $m$ - mass of moving particle
$M -$ Mass of particle at rest
$\Rightarrow K . E_{T h} = 10.2 e V [1 + \frac{1}{1}] = 20.4 e V$

PHXII12:ATOMS

356378 A hydrogen atom moves with a velocity $u$ and makes a head on inelastic collision are in the ground state before collision. What is the minimum value of $u$ , if one of them is to be given a minimum excitation energy? The ionization energy is $13.6 e V$ . Mass of the hydrogen atom is $1.0078 \times 1.66 \times 10^{- 27} k g$ .

1

3.74 \times 10^{4} m / s

2

7.35 \times 10^{4} m / s

3

1.64 \times 10^{4} m / s

4

6.24 \times 10^{4} m / s

Explanation:

From conservation of momentum $m u = 2 m v$
$\Rightarrow v = u / 2 (1)$
The energy of excitation $Δ E$ is given by
$Δ E = \frac{1}{2} m u^{2} - 2 \times \frac{1}{2} m (u / 2)^{2}$
$\Rightarrow Δ E = \frac{1}{4} m u^{2} (2)$
The minimum excitation energy for a $H$ -atom is for transition $n_{1} = 1$ to $n_{2} = 2$ which corresponds to an energy of $10.2 e V$ . From Eq. (2)
$\frac{1}{4} \times (1.0078 \times 1.66 \times 10^{- 27}) u^{2}$
$= 10.2 \times (1.6 \times 10^{- 19})$
$\Rightarrow u = 6.24 \times 10^{4} m / s$

PHXII12:ATOMS

356379 A $H e^{+}$ ion is at rest and is in ground state. A neutron with intial kinetic energy $K$ collides head on with the $H e^{+}$ ion. Find minimum value of $K$ so that there can be an inelastic collision between these two particles.
supporting img

1

51 e V

2

30.6 e V

3

20.4 e V

4

40.8 e V

Explanation:

Method-1
From conservation of linear momentum
$m u = (m + 4 m) v \Rightarrow v = \frac{u}{5};$
In this case loss in $K E$ will be maximum. Now this $Δ K_{max}$ must be greater than the energy received to excite
$| Δ K_{max} | = \frac{1}{2} m u^{2} - \frac{1}{2} (5 m) {(\frac{u}{5})}^{2} = \frac{1}{2} m u^{2} (1 - \frac{1}{5})$
$= \frac{1}{2} m u^{2} \cdot \frac{4}{5}$
$∴ \frac{1}{2} m u^{2} \cdot \frac{4}{5} \geq 40.8 \Rightarrow \frac{1}{2} m u^{2} \geq 51 e V$
Hence, minimum value of $K$ for an inelastic collision: $K_{min} = 51 e V .$
Method-2
The Thereshold energy
$K . E_{T h} = E_{a} [1 + \frac{m}{M}] (M \approx 4 m)$
$E_{1} = \frac{- 13.6 \times 2^{2}}{1^{2}}, E_{2} = \frac{- 13.6 \times 2^{2}}{2^{2}}$
$\Rightarrow E_{a} = E_{1} - E_{2}$
$\Rightarrow E_{a} = 13.6 \times 3 e V \Rightarrow K . E_{T h} = \frac{5 E_{a}}{4} = 51 e V$

PHXII12:ATOMS

356380 Consider a hydrogen atom present at rest in ground state. A fast moving neutron collides with the atom. Find the threshold energy of the neutron for excitations $n_{1} = 2 t o n_{2} = 3$ $(m_{N} ≃ m_{H})$

1

24.2 e V

2

12.4 e V

3

10.2 e V

4

20.4 e V

Explanation:

The threshold energy
$E_{T h} = E_{a} [1 + \frac{m}{M}]$
$m = m_{n}$ & $M = m_{H}$
$E_{a} = 10.2 e V & m_{n} \approx m_{H}$
In the following table threshold energies are given for different excitations of $H$ -atom
It can be observed from the above table that $E_{T h} = 24.2 e V$
supporting img

PHXII12:ATOMS

356376 An electron with $K E 6 e V$ is incident on a hydrogen atom in its ground state. The collision

1 May be partially inelastic

2 Must be completely inelastic

3 May be partially elastic

4 Must be elastic

Explanation:

The minimum energy required for excitation of hydrogen atom is 10.2 $e V$ . Here the $K$ . $E$ of the incoming electron is less than 10.2 $e V$ . So excitation does not take place. The $K$ . $E$ of the electron will be retained as $K$ . $E$ for the system. The collision must be perfectly elastic.

PHXII12:ATOMS

356377 A moving hydrogen atom makes a head-on collision with a stationary hydrogen atom. Before collision, both atoms are in ground state and after collision they move together. What is the minimum value of the kinetic energy of the moving hydrogen atom, such that one of the atoms reaches one of the excitation state?
supporting img

1

20.4 e V

2

10.2 e V

3

30.6 e V

4

A n y v a l u e

Explanation:

Method-1
From conservation of linear momentum
$m u + 0 = (m + m) v \Rightarrow v = \frac{u}{2};$
In this case loss in $K E$ will be maximum. Now this $Δ K_{max}$ must be greater than the energy required to excite
$| {(Δ K)}_{max} | = | \frac{1}{2} 2 m {(\frac{u}{2})}^{2} - \frac{1}{2} m u^{2} | = \frac{1}{4} m u^{2}$
$\frac{1}{4} m u^{2} \geq 10.2 e v \Rightarrow \frac{1}{2} m u^{2} \geq 20.4 e V$
Therefore, the minimum kinetic energy of moving hydrogen is $20.4 e V$ .
Method-2
The threshold energy ( $K . E_{min}$ ) required for the moving $H$ -atom is
$K . E_{T h} = E_{a} [1 + \frac{m}{M}]$
$E_{a} -$ Energy of excitation, $m$ - mass of moving particle
$M -$ Mass of particle at rest
$\Rightarrow K . E_{T h} = 10.2 e V [1 + \frac{1}{1}] = 20.4 e V$

PHXII12:ATOMS

356378 A hydrogen atom moves with a velocity $u$ and makes a head on inelastic collision are in the ground state before collision. What is the minimum value of $u$ , if one of them is to be given a minimum excitation energy? The ionization energy is $13.6 e V$ . Mass of the hydrogen atom is $1.0078 \times 1.66 \times 10^{- 27} k g$ .

1

3.74 \times 10^{4} m / s

2

7.35 \times 10^{4} m / s

3

1.64 \times 10^{4} m / s

4

6.24 \times 10^{4} m / s

Explanation:

From conservation of momentum $m u = 2 m v$
$\Rightarrow v = u / 2 (1)$
The energy of excitation $Δ E$ is given by
$Δ E = \frac{1}{2} m u^{2} - 2 \times \frac{1}{2} m (u / 2)^{2}$
$\Rightarrow Δ E = \frac{1}{4} m u^{2} (2)$
The minimum excitation energy for a $H$ -atom is for transition $n_{1} = 1$ to $n_{2} = 2$ which corresponds to an energy of $10.2 e V$ . From Eq. (2)
$\frac{1}{4} \times (1.0078 \times 1.66 \times 10^{- 27}) u^{2}$
$= 10.2 \times (1.6 \times 10^{- 19})$
$\Rightarrow u = 6.24 \times 10^{4} m / s$

PHXII12:ATOMS

356379 A $H e^{+}$ ion is at rest and is in ground state. A neutron with intial kinetic energy $K$ collides head on with the $H e^{+}$ ion. Find minimum value of $K$ so that there can be an inelastic collision between these two particles.
supporting img

1

51 e V

2

30.6 e V

3

20.4 e V

4

40.8 e V

Explanation:

Method-1
From conservation of linear momentum
$m u = (m + 4 m) v \Rightarrow v = \frac{u}{5};$
In this case loss in $K E$ will be maximum. Now this $Δ K_{max}$ must be greater than the energy received to excite
$| Δ K_{max} | = \frac{1}{2} m u^{2} - \frac{1}{2} (5 m) {(\frac{u}{5})}^{2} = \frac{1}{2} m u^{2} (1 - \frac{1}{5})$
$= \frac{1}{2} m u^{2} \cdot \frac{4}{5}$
$∴ \frac{1}{2} m u^{2} \cdot \frac{4}{5} \geq 40.8 \Rightarrow \frac{1}{2} m u^{2} \geq 51 e V$
Hence, minimum value of $K$ for an inelastic collision: $K_{min} = 51 e V .$
Method-2
The Thereshold energy
$K . E_{T h} = E_{a} [1 + \frac{m}{M}] (M \approx 4 m)$
$E_{1} = \frac{- 13.6 \times 2^{2}}{1^{2}}, E_{2} = \frac{- 13.6 \times 2^{2}}{2^{2}}$
$\Rightarrow E_{a} = E_{1} - E_{2}$
$\Rightarrow E_{a} = 13.6 \times 3 e V \Rightarrow K . E_{T h} = \frac{5 E_{a}}{4} = 51 e V$

PHXII12:ATOMS

356380 Consider a hydrogen atom present at rest in ground state. A fast moving neutron collides with the atom. Find the threshold energy of the neutron for excitations $n_{1} = 2 t o n_{2} = 3$ $(m_{N} ≃ m_{H})$

1

24.2 e V

2

12.4 e V

3

10.2 e V

4

20.4 e V

Explanation:

The threshold energy
$E_{T h} = E_{a} [1 + \frac{m}{M}]$
$m = m_{n}$ & $M = m_{H}$
$E_{a} = 10.2 e V & m_{n} \approx m_{H}$
In the following table threshold energies are given for different excitations of $H$ -atom
It can be observed from the above table that $E_{T h} = 24.2 e V$
supporting img

PHXII12:ATOMS

356376 An electron with $K E 6 e V$ is incident on a hydrogen atom in its ground state. The collision

1 May be partially inelastic

2 Must be completely inelastic

3 May be partially elastic

4 Must be elastic

Explanation:

The minimum energy required for excitation of hydrogen atom is 10.2 $e V$ . Here the $K$ . $E$ of the incoming electron is less than 10.2 $e V$ . So excitation does not take place. The $K$ . $E$ of the electron will be retained as $K$ . $E$ for the system. The collision must be perfectly elastic.

PHXII12:ATOMS

356377 A moving hydrogen atom makes a head-on collision with a stationary hydrogen atom. Before collision, both atoms are in ground state and after collision they move together. What is the minimum value of the kinetic energy of the moving hydrogen atom, such that one of the atoms reaches one of the excitation state?
supporting img

1

20.4 e V

2

10.2 e V

3

30.6 e V

4

A n y v a l u e

Explanation:

Method-1
From conservation of linear momentum
$m u + 0 = (m + m) v \Rightarrow v = \frac{u}{2};$
In this case loss in $K E$ will be maximum. Now this $Δ K_{max}$ must be greater than the energy required to excite
$| {(Δ K)}_{max} | = | \frac{1}{2} 2 m {(\frac{u}{2})}^{2} - \frac{1}{2} m u^{2} | = \frac{1}{4} m u^{2}$
$\frac{1}{4} m u^{2} \geq 10.2 e v \Rightarrow \frac{1}{2} m u^{2} \geq 20.4 e V$
Therefore, the minimum kinetic energy of moving hydrogen is $20.4 e V$ .
Method-2
The threshold energy ( $K . E_{min}$ ) required for the moving $H$ -atom is
$K . E_{T h} = E_{a} [1 + \frac{m}{M}]$
$E_{a} -$ Energy of excitation, $m$ - mass of moving particle
$M -$ Mass of particle at rest
$\Rightarrow K . E_{T h} = 10.2 e V [1 + \frac{1}{1}] = 20.4 e V$

PHXII12:ATOMS

356378 A hydrogen atom moves with a velocity $u$ and makes a head on inelastic collision are in the ground state before collision. What is the minimum value of $u$ , if one of them is to be given a minimum excitation energy? The ionization energy is $13.6 e V$ . Mass of the hydrogen atom is $1.0078 \times 1.66 \times 10^{- 27} k g$ .

1

3.74 \times 10^{4} m / s

2

7.35 \times 10^{4} m / s

3

1.64 \times 10^{4} m / s

4

6.24 \times 10^{4} m / s

Explanation:

From conservation of momentum $m u = 2 m v$
$\Rightarrow v = u / 2 (1)$
The energy of excitation $Δ E$ is given by
$Δ E = \frac{1}{2} m u^{2} - 2 \times \frac{1}{2} m (u / 2)^{2}$
$\Rightarrow Δ E = \frac{1}{4} m u^{2} (2)$
The minimum excitation energy for a $H$ -atom is for transition $n_{1} = 1$ to $n_{2} = 2$ which corresponds to an energy of $10.2 e V$ . From Eq. (2)
$\frac{1}{4} \times (1.0078 \times 1.66 \times 10^{- 27}) u^{2}$
$= 10.2 \times (1.6 \times 10^{- 19})$
$\Rightarrow u = 6.24 \times 10^{4} m / s$

PHXII12:ATOMS

356379 A $H e^{+}$ ion is at rest and is in ground state. A neutron with intial kinetic energy $K$ collides head on with the $H e^{+}$ ion. Find minimum value of $K$ so that there can be an inelastic collision between these two particles.
supporting img

1

51 e V

2

30.6 e V

3

20.4 e V

4

40.8 e V

Explanation:

Method-1
From conservation of linear momentum
$m u = (m + 4 m) v \Rightarrow v = \frac{u}{5};$
In this case loss in $K E$ will be maximum. Now this $Δ K_{max}$ must be greater than the energy received to excite
$| Δ K_{max} | = \frac{1}{2} m u^{2} - \frac{1}{2} (5 m) {(\frac{u}{5})}^{2} = \frac{1}{2} m u^{2} (1 - \frac{1}{5})$
$= \frac{1}{2} m u^{2} \cdot \frac{4}{5}$
$∴ \frac{1}{2} m u^{2} \cdot \frac{4}{5} \geq 40.8 \Rightarrow \frac{1}{2} m u^{2} \geq 51 e V$
Hence, minimum value of $K$ for an inelastic collision: $K_{min} = 51 e V .$
Method-2
The Thereshold energy
$K . E_{T h} = E_{a} [1 + \frac{m}{M}] (M \approx 4 m)$
$E_{1} = \frac{- 13.6 \times 2^{2}}{1^{2}}, E_{2} = \frac{- 13.6 \times 2^{2}}{2^{2}}$
$\Rightarrow E_{a} = E_{1} - E_{2}$
$\Rightarrow E_{a} = 13.6 \times 3 e V \Rightarrow K . E_{T h} = \frac{5 E_{a}}{4} = 51 e V$

PHXII12:ATOMS

356380 Consider a hydrogen atom present at rest in ground state. A fast moving neutron collides with the atom. Find the threshold energy of the neutron for excitations $n_{1} = 2 t o n_{2} = 3$ $(m_{N} ≃ m_{H})$

1

24.2 e V

2

12.4 e V

3

10.2 e V

4

20.4 e V

Explanation:

The threshold energy
$E_{T h} = E_{a} [1 + \frac{m}{M}]$
$m = m_{n}$ & $M = m_{H}$
$E_{a} = 10.2 e V & m_{n} \approx m_{H}$
In the following table threshold energies are given for different excitations of $H$ -atom
It can be observed from the above table that $E_{T h} = 24.2 e V$
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