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PHXII12:ATOMS

356353 In Rutherford scattering experiment, what will be the correct angle for $α$ scattering for an impact parameter $b = 0$ ?

1

90^{\circ}

2

270^{\circ}

3

0^{\circ}

4

180^{\circ}

Explanation:

We know that impact parameter,
$b α \cot \frac{θ}{2} \Rightarrow b = K \cot \frac{θ}{2}$
Here, $K$ is constant $b = 0$ . Hence, $θ = 180^{\circ}$

PHXII12:ATOMS

356354 In a head on collision between an $α$ -particle and gold nucleus, the closest distance of approach is $41.3 f m$ . Calculate the energy of $α$ -particle. $(z$ of gold $= 79)$

1

5.51 e V

2

5.51 M e V

3

5.51 J

4

5.51 K J

Explanation:

$E = \frac{(Z e) (2 e)}{4 π ε_{0} r} = \frac{(\begin{array}{l} 9 \times 10^{9} \times 79 \times \\ 2 {(1.6 \times 10^{- 19})}^{2} \end{array})}{41.3 \times 10^{- 15}}$
$= \frac{9 \times 79 \times 3.2 \times 1.6 \times 10^{- 14}}{41.3} J$
$= 8.814 \times 10^{- 13} J o u l e = \frac{8.814 \times 10^{- 13}}{1.6 \times 10^{- 13}}$
$E = 5.51 M e V .$

PHXII12:ATOMS

356355 In the Rutherford's alpha scattering experiment, as the impact parameter increases, the scattering angle of the alpha particle

1 decreases

2 increases

3 remains the same

4 is always

90^{\circ}

Explanation:

Impact parameter $b$ can be written as:
$b = \frac{1}{4 π ε_{0}} \frac{Z e^{2} \cot (θ / 2)}{(\frac{1}{2} m υ^{2})}$
supporting img
If impact meter increases then scattering angle decreases. Correct option is (1).

KCET - 2023

PHXII12:ATOMS

356356 In scattering experiment the force that scatters particles is

1 Nuclear force

2 Coulomb’s force

3 Both(1)and(2)

4 Gravitational force

PHXII12:ATOMS

356353 In Rutherford scattering experiment, what will be the correct angle for $α$ scattering for an impact parameter $b = 0$ ?

1

90^{\circ}

2

270^{\circ}

3

0^{\circ}

4

180^{\circ}

Explanation:

We know that impact parameter,
$b α \cot \frac{θ}{2} \Rightarrow b = K \cot \frac{θ}{2}$
Here, $K$ is constant $b = 0$ . Hence, $θ = 180^{\circ}$

PHXII12:ATOMS

356354 In a head on collision between an $α$ -particle and gold nucleus, the closest distance of approach is $41.3 f m$ . Calculate the energy of $α$ -particle. $(z$ of gold $= 79)$

1

5.51 e V

2

5.51 M e V

3

5.51 J

4

5.51 K J

Explanation:

$E = \frac{(Z e) (2 e)}{4 π ε_{0} r} = \frac{(\begin{array}{l} 9 \times 10^{9} \times 79 \times \\ 2 {(1.6 \times 10^{- 19})}^{2} \end{array})}{41.3 \times 10^{- 15}}$
$= \frac{9 \times 79 \times 3.2 \times 1.6 \times 10^{- 14}}{41.3} J$
$= 8.814 \times 10^{- 13} J o u l e = \frac{8.814 \times 10^{- 13}}{1.6 \times 10^{- 13}}$
$E = 5.51 M e V .$

PHXII12:ATOMS

356355 In the Rutherford's alpha scattering experiment, as the impact parameter increases, the scattering angle of the alpha particle

1 decreases

2 increases

3 remains the same

4 is always

90^{\circ}

Explanation:

Impact parameter $b$ can be written as:
$b = \frac{1}{4 π ε_{0}} \frac{Z e^{2} \cot (θ / 2)}{(\frac{1}{2} m υ^{2})}$
supporting img
If impact meter increases then scattering angle decreases. Correct option is (1).

KCET - 2023

PHXII12:ATOMS

356356 In scattering experiment the force that scatters particles is

1 Nuclear force

2 Coulomb’s force

3 Both(1)and(2)

4 Gravitational force

PHXII12:ATOMS

356353 In Rutherford scattering experiment, what will be the correct angle for $α$ scattering for an impact parameter $b = 0$ ?

1

90^{\circ}

2

270^{\circ}

3

0^{\circ}

4

180^{\circ}

Explanation:

We know that impact parameter,
$b α \cot \frac{θ}{2} \Rightarrow b = K \cot \frac{θ}{2}$
Here, $K$ is constant $b = 0$ . Hence, $θ = 180^{\circ}$

PHXII12:ATOMS

356354 In a head on collision between an $α$ -particle and gold nucleus, the closest distance of approach is $41.3 f m$ . Calculate the energy of $α$ -particle. $(z$ of gold $= 79)$

1

5.51 e V

2

5.51 M e V

3

5.51 J

4

5.51 K J

Explanation:

$E = \frac{(Z e) (2 e)}{4 π ε_{0} r} = \frac{(\begin{array}{l} 9 \times 10^{9} \times 79 \times \\ 2 {(1.6 \times 10^{- 19})}^{2} \end{array})}{41.3 \times 10^{- 15}}$
$= \frac{9 \times 79 \times 3.2 \times 1.6 \times 10^{- 14}}{41.3} J$
$= 8.814 \times 10^{- 13} J o u l e = \frac{8.814 \times 10^{- 13}}{1.6 \times 10^{- 13}}$
$E = 5.51 M e V .$

PHXII12:ATOMS

356355 In the Rutherford's alpha scattering experiment, as the impact parameter increases, the scattering angle of the alpha particle

1 decreases

2 increases

3 remains the same

4 is always

90^{\circ}

Explanation:

Impact parameter $b$ can be written as:
$b = \frac{1}{4 π ε_{0}} \frac{Z e^{2} \cot (θ / 2)}{(\frac{1}{2} m υ^{2})}$
supporting img
If impact meter increases then scattering angle decreases. Correct option is (1).

KCET - 2023

PHXII12:ATOMS

356356 In scattering experiment the force that scatters particles is

1 Nuclear force

2 Coulomb’s force

3 Both(1)and(2)

4 Gravitational force

PHXII12:ATOMS

356353 In Rutherford scattering experiment, what will be the correct angle for $α$ scattering for an impact parameter $b = 0$ ?

1

90^{\circ}

2

270^{\circ}

3

0^{\circ}

4

180^{\circ}

Explanation:

We know that impact parameter,
$b α \cot \frac{θ}{2} \Rightarrow b = K \cot \frac{θ}{2}$
Here, $K$ is constant $b = 0$ . Hence, $θ = 180^{\circ}$

PHXII12:ATOMS

356354 In a head on collision between an $α$ -particle and gold nucleus, the closest distance of approach is $41.3 f m$ . Calculate the energy of $α$ -particle. $(z$ of gold $= 79)$

1

5.51 e V

2

5.51 M e V

3

5.51 J

4

5.51 K J

Explanation:

$E = \frac{(Z e) (2 e)}{4 π ε_{0} r} = \frac{(\begin{array}{l} 9 \times 10^{9} \times 79 \times \\ 2 {(1.6 \times 10^{- 19})}^{2} \end{array})}{41.3 \times 10^{- 15}}$
$= \frac{9 \times 79 \times 3.2 \times 1.6 \times 10^{- 14}}{41.3} J$
$= 8.814 \times 10^{- 13} J o u l e = \frac{8.814 \times 10^{- 13}}{1.6 \times 10^{- 13}}$
$E = 5.51 M e V .$

PHXII12:ATOMS

356355 In the Rutherford's alpha scattering experiment, as the impact parameter increases, the scattering angle of the alpha particle

1 decreases

2 increases

3 remains the same

4 is always

90^{\circ}

Explanation:

Impact parameter $b$ can be written as:
$b = \frac{1}{4 π ε_{0}} \frac{Z e^{2} \cot (θ / 2)}{(\frac{1}{2} m υ^{2})}$
supporting img
If impact meter increases then scattering angle decreases. Correct option is (1).

KCET - 2023

PHXII12:ATOMS

356356 In scattering experiment the force that scatters particles is

1 Nuclear force

2 Coulomb’s force

3 Both(1)and(2)

4 Gravitational force

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