Question

An α-particle has a charge of +2e and a mass of 6.64 10-27 kg. It is accelerated from rest through a potential difference that has a value of 1.24 106 V and then enters a uniform magnetic field whose magnitude is 2.50 T. The α-particle moves perpendicular to the magnetic field at all times.

Answer 1

1.1 x 10^7 m/s

Step-by-step explanation:

mass of electron, m = 6.64 x 10^-27 kg

charge of electron, q = 2 x 1.6 x 10^-19 C

potential difference, V = 1.24 x 10^6 V

Magnetic field, B = 2.5 T

Let v be the velocity of alpha particle

Kinetic energy is given by q V

0.5 x m v² = q V

0.5 x 6.64 x 10^-27 x v² = 3.2 x 10^-19 x 1.24 x 10^6

3.32 x 10^-27 x v² = 3.968 x 10^-13

v² = 1.195 x 10^14

v = 1.1 x 10^7 m/s

Thus, the speed of alpha particles is 1.1 x 10^7 m/s.

Answer 2

`v_2=1.09* 10^7\ m/s`

Step-by-step explanation:

It is given that,

Charge of alpha particle,
`q=2e=2* 1.6* 10^(-19)=3.2* 10^(-19)\ C`

Mass of the alpha particle,
`m=6.64* 10^(-27)\ kg`

Potential difference,
`(V_1-V_2)=1.24* 10^6\ V`

Magnetic field, B = 2.5 T

The α-particle moves perpendicular to the magnetic field at all times. The initial speed of the alpha particle is 0 as it is at rest. Using the conservation of energy as :

`(1)/(2)mv_1^2+qV_1=(1)/(2)mv_2^2+qV_2`

`v_2` is the speed of the α–particle

`q(V_1-V_2)=(1)/(2)mv_2^2`

`v_2=\sqrt{(2q(V_1-V_2))/(m)}`

`v_2=\sqrt{(2* 3.2* 10^(-19)* 1.24* 10^6)/(6.64* 10^(-27))}`

`v_2=1.09* 10^7\ m/s`

So, the speed of alpha particle is
`1.09* 10^7\ m/s`. Hence, this is the required solution.