Question

An alpha particle (q = +2e, m = 4.00 u) travels in a circular path of radius 5.56 cm in a uniform magnetic field with B = 1.76 T. Calculate (a) its speed, (b) its period of revolution, (c) its kinetic energy, and (d)the potential difference through which it would have to be accelerated to achieve this energy.

I just need part c and d

Answer 1

Given :-

• 
  q = +20
     : Charge of alpha particle
• 
  m = 4.00 u
                       : Mass of alpha particle.
• 
  r = 5.56 cm = 0.0556 m
  : Radius of circular path
• 
  B=1.767
                         : Magnetic field

Let,

• $\small v$    : be the velocity of the alpha particle
• $\small K$ : be the kinetic energy of the alpha particle
• 
  c=1.6 x 10-19
  C : be the charge
• $\small \Delta V$   : be the potential difference.
•    : be the period.

Mass m and charge q of the alpha particle in SI unit is,

• 
  m = 4u = 4 x 1.66 x 10-27 = 6.64 x 10-27 kg
  
• 
  q=+2e = 2 x 1.6 x 10-19 = 3.2 x 10-19 J
  

(a) its speed :-

Velocity of a charged particle, in magnetic field moving in circular path, is given as,

$v =\frac{qrB}{m}$

$v =\frac{3.2 \times 10^{-19} \times 0.0556 \times 1.76}{ 6.64 \times 10^{-27}}$

v = 4.72 x 106 m/s

(b) Its period of revolution :-

Period of a charged particle in magnetic field is given as,

$T = \frac{2 \pi r}{v}$

T = 2 x 3.141 x 0.0556 4.72 x 106

$\mathbf{T = 0.74 \times 10 ^{-7} \: s}$

(c) Its kinetic energy :-

Kinetic energy is given as,

$K =\frac{1}{2} mv^{2}$

$K =\frac{1}{2}\times 6.64 \times 10^{-27} \times \left ( 4.72 \times10^{6} \right )^{2}$

$K =3.32 \times 10^{-27} \times 22.28 \times10^{12}$

$K =73.96 \times 10^{-15} \: J$

in eV unit,

$\mathbf{K = 4.62 \times 10^{5} \: eV}$

This is the kinetic energy of the alpha particle.

(d) The potential difference :-

The potential difference is given as,

$\Delta V = \frac{K}{q}$

$\Delta V = \frac{4.62 \times10^{5} \: \left (eV \right )}{2\left (e \right )}$

$\mathbf{ \Delta V = 2.31 \times10^{5} \: V }$

This is the potential difference through which it would have to be accelerated to achieve this energy.

Answer 2

Given :-

• 
  q = +20
     : Charge of alpha particle
• 
  m = 4.00 u
                       : Mass of alpha particle.
• 
  r = 5.56 cm = 0.0556 m
  : Radius of circular path
• 
  B=1.767
                         : Magnetic field

Let,

• $\small v$    : be the velocity of the alpha particle
• $\small K$ : be the kinetic energy of the alpha particle
• 
  c=1.6 x 10-19
  C : be the charge
• $\small \Delta V$   : be the potential difference.
•    : be the period.

Mass m and charge q of the alpha particle in SI unit is,

• 
  m = 4u = 4 x 1.66 x 10-27 = 6.64 x 10-27 kg
  
• 
  q=+2e = 2 x 1.6 x 10-19 = 3.2 x 10-19 J
  

(a) its speed :-

Velocity of a charged particle, in magnetic field moving in circular path, is given as,

$v =\frac{qrB}{m}$

$v =\frac{3.2 \times 10^{-19} \times 0.0556 \times 1.76}{ 6.64 \times 10^{-27}}$

v = 4.72 x 106 m/s

(b) Its period of revolution :-

Period of a charged particle in magnetic field is given as,

$T = \frac{2 \pi r}{v}$

T = 2 x 3.141 x 0.0556 4.72 x 106

$\mathbf{T = 0.74 \times 10 ^{-7} \: s}$

(c) Its kinetic energy :-

Kinetic energy is given as,

$K =\frac{1}{2} mv^{2}$

$K =\frac{1}{2}\times 6.64 \times 10^{-27} \times \left ( 4.72 \times10^{6} \right )^{2}$

$K =3.32 \times 10^{-27} \times 22.28 \times10^{12}$

$K =73.96 \times 10^{-15} \: J$

in eV unit,

$\mathbf{K = 4.62 \times 10^{5} \: eV}$

This is the kinetic energy of the alpha particle.

(d) The potential difference :-

The potential difference is given as,

$\Delta V = \frac{K}{q}$

$\Delta V = \frac{4.62 \times10^{5} \: \left (eV \right )}{2\left (e \right )}$

$\mathbf{ \Delta V = 2.31 \times10^{5} \: V }$

This is the potential difference through which it would have to be accelerated to achieve this energy.