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

Answer:

$v_2=1.09\times 10^7\ m/s$

Explanation:

It is given that,

Charge of alpha particle, $q=2e=2\times 1.6\times 10^{-19}=3.2\times 10^{-19}\ C$

Mass of the alpha particle, $m=6.64\times 10^{-27}\ kg$

Potential difference, $(V_1-V_2)=1.24\times 10^6\ V$

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 :

$\dfrac{1}{2}mv_1^2+qV_1=\dfrac{1}{2}mv_2^2+qV_2$

$v_2$ is the speed of the  α–particle

$q(V_1-V_2)=\dfrac{1}{2}mv_2^2$

$v_2=\sqrt{\dfrac{2q(V_1-V_2)}{m}}$

$v_2=\sqrt{\dfrac{2\times 3.2\times 10^{-19}\times 1.24\times 10^6}{6.64\times 10^{-27}}}$

$v_2=1.09\times 10^7\ m/s$

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

Answer 2

Answer:

1.1 x 10^7 m/s

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.