Question

An alpha particle (the nucleus of helium
atom) (with charge +2) is accelerated
and moves in a vacuum tube with kinetic
energy = 10 MeV. It passes through a
uniform magnetic
field of 1.88 T
and traces a circular path of
ratius 24.6 cm. obtain the mass
of
the alpha particle. ( charge on electron= 1.6×10^-19C

With explanation detailed

Please no one liners

Answer 1

Given:

An alpha particle (the nucleus of helium atom) (with charge +2) is accelerated and moves in a vacuum tube with kinetic energy = 10 MeV. It passes through a uniform magnetic field of 1.88 T and traces a circular path of radius 24.6 cm.

To find:

Mass of particle ?

Calculation:

The radius of a charged particle in magnetic field is given as :

$\therefore \: r = \dfrac{m \times v}{q \times B }$

$\implies \: r = \dfrac{ \sqrt{2 \times m \times KE} }{q \times B }$

$\implies \: {r}^{2} = \dfrac{2 \times m \times KE }{{(q \times B)}^{2} }$

$\implies \: m = \dfrac{{q}^{2} \times {B}^{2} \times {r}^{2}}{2 \times KE}$

$\implies \: m = \dfrac{{(2 \times 1.6 \times {10}^{ - 19} )}^{2} \times {(1.88)}^{2} \times {(0.246)}^{2}}{2 \times (10 \times {10}^{6} \times 1.6 \times {10}^{ - 19} ) }$

$\implies \: m = \dfrac{4 \times 1.6 \times {10}^{ - 19} \times 3.35 \times 0.06}{2 \times 10 \times {10}^{6} }$

$\implies \: m = \dfrac{4 \times 1.6 \times {10}^{ - 21} \times 3.35 \times 6}{2 \times {10}^{7} }$

$\implies \: m = 64.32 \times {10}^{ - 28} \: kg$

So, mass of alpha particle is 64.32 × 10^(-28) kg.

Answer 2

Answer is "do it by yourself, try hard you'll find the answer...and if you find out then tell me also...oki..bye