Question

An alpha-particle and a proton of the same kinetic energy are in turn allowed to pass through a magnetic field B, acting normal to the direction of motion of the particles. Calculate the ratio of radii of the circular paths described by them.

Answer 1

answer : ratio of radii of circular paths described by them is 1 : 1

explanation : we know, mass of proton, $m_p$= 1/4 × mass of alpha particle , $m_{\alpha}$

charge on proton, $q_p$ =1/2 × charge on alpha particle , $q_{\alpha}$

using formula, $r=\frac{mv}{qB}$

we know, mv = √{2Km}, where k is kinetic energy.

so, r = √{2Km}/qB

here, K, v and B are constant terms

so, $r\propto\frac{\sqrt{m}}{q}$

so, relation will be ...

$\bf{\frac{r_p}{r_{\alpha}}=\sqrt{\frac{m_p}{m_{\alpha}}}\frac{q_{\alpha}}{q_p}}$

= $\bf{\sqrt{\frac{m_p}{4m_p}}\frac{2q_p}{q_p} }$

= $\frac{1}{1}$

hence, ratio of their radii of the circular path described by the particles is 1 : 1.

Answer 2

Answer:

answer : ratio of radii of circular paths described by them is 1 : 1

explanation : we know, mass of proton, m_pm

p

= 1/4 × mass of alpha particle , m_{\alpha}m

α

charge on proton, q_pq

p

=1/2 × charge on alpha particle , q_{\alpha}q

α

using formula, r=\frac{mv}{qB}r=

qB

mv

we know, mv = √{2Km}, where k is kinetic energy.

so, r = √{2Km}/qB

here, K, v and B are constant terms

so, r\propto\frac{\sqrt{m}}{q}r∝

q

m

so, relation will be ...

\bf{\frac{r_p}{r_{\alpha}}=\sqrt{\frac{m_p}{m_{\alpha}}}\frac{q_{\alpha}}{q_p}}

r

α

r

p

=

m

α

m

p

q

p

q

α

= \bf{\sqrt{\frac{m_p}{4m_p}}\frac{2q_p}{q_p} }

4m

p

m

p

q

p

2q

p

= \frac{1}{1}

1

1

hence, ratio of their radii of the circular path described by the particles is 1 : 1.