Question

Hi friends !

Solve this , with process . ​

Answer 1

we know that

r=mv/qb

r=p/qb

where p =mv=momentum

also we know

momentum=√2mke

so we can write

r=√2mKe/qb

now we know that

Ke and b is same for all the three subatomic particles so,

we can write

r directly proportional to (√m)/q

re=√Me/e......i)

rp=√Mp/e......ii)

mass of alpha particle is 4 times of proton mass and have charge 2

ra=√4Mp/2e

ra=√Mp/e.......III)

Me<Mp.....iv)

comparing all the above eqaution we can write

re<rp=ra

Answer 2

The kinetic energy of a particle revolving around a circular path in a uniform magnetic field is given by,

$\sf{\longrightarrow K=\dfrac{q^2B^2r^2}{2m}}$

where

• $\sf{q=}$ charge of the particle

• $\sf{B=}$ magnetic field

• $\sf{r=}$ radius of the circular path

• $\sf{m=}$ mass of the particle

The equation can be rewritten as,

$\sf{\longrightarrow 2mK=q^2B^2r^2}$

In the question $\sf{K}$ and $\sf{B}$ are constants. So,

$\sf{\longrightarrow m\propto q^2r^2}$

$\sf{\longrightarrow r\propto\dfrac{\sqrt m}{q}}$

Let,

$\sf{\longrightarrow r=k\cdot\dfrac{\sqrt m}{q}}$

where $\sf{k}$ is a positive constant.

Mass of electron $\sf{=9.1\times10^{-31}\ kg.}$

Charge of electron $\sf{=1.6\times10^{-19}\ C.}$

Then, radius of path of electron,

$\sf{\longrightarrow r_e=k\cdot\dfrac{\sqrt{9.1\times10^{-31}}}{1.6\times10^{-19}}}$

$\sf{\longrightarrow r_e=6.0k\times10^3\ m}$

Mass of proton $\sf{=1.7\times10^{-27}\ kg.}$

Charge of proton $\sf{=1.6\times10^{-19}\ C.}$

Then, radius of path of proton,

$\sf{\longrightarrow r_p=k\cdot\dfrac{\sqrt{1.7\times10^{-27}}}{1.6\times10^{-19}}}$

$\sf{\longrightarrow r_p=2.6k\times10^5\ m}$

Mass of alpha particle $\sf{=6.8\times10^{-27}\ kg.}$

Charge of alpha particle $\sf{=3.2\times10^{-19}\ C.}$

Then, radius of path of proton,

$\sf{\longrightarrow r_\alpha=k\cdot\dfrac{\sqrt{6.8\times10^{-27}}}{3.2\times10^{-19}}}$

$\sf{\longrightarrow r_\alpha=2.6k\times10^5\ m}$

Comparing each radius we get,

$\sf{\longrightarrow\underline{\underline{r_e<r_p=r_\alpha}}}$

Hence (D) is the answer.