The distance of closest approach of an alpha-particle fired towards a nucleus with momentum p is r. If the momentum of the alpha-particle is 2p, the corresponding distance of closest approach is :

Question

The distance of closest approach of an alpha- particle fired towards a nucleus with momentum p-is r- The distance of closest approach when the momentum of alpha-particle is 2p is2rr-44rr-2

Toppr Admin · Accepted Answer

12mv2-xA0-q1-xA0-q24-x3C0-x3F5-0-xA0-rInitially- -xA0-p22m-2e-xA0-q4-x3C0-x3F5-0-xA0-r -xA0- -xA0- -1-Now- -xA0-2p-22m-2eq4-x3C0-x3F5-0-xA0-r1 -xA0- -xA0- -II-Equation -xA0-I-xF7-ii-14-r1rr1-r4

The distance of closest approach of an $α$ - particle fired towards a nucleus with momentum $p$ ,is $r$ . The distance of closest approach when the momentum of $α$ -particle is $2 p$ is
$2 r$
$r / 4$
$4 r$
$r / 2$

$\frac{1}{2} m v^{2} = \frac{q_{1} q_{2}}{4 π ϵ_{0 r}}$

Initially, $\frac{p^{2}}{2 m} = \frac{2 e q}{4 π ϵ_{0} r}$ ----(1)

Now, $\frac{(2 p)^{2}}{2 m} = \frac{2 e q}{4 π ϵ_{0} r^{1}}$ ----(II)

Equation $(I) \div (i i)$

$\frac{1}{4} = \frac{r^{1}}{r}$

$r^{1} = \frac{r}{4}$

The distance of closest approach of an α- particle fired towards a nucleus with momentum p,is r. The distance of closest approach when the momentum of α-particle is 2p is2rr/44rr/2

12mv2= q1 q24πϵ0 rInitially, p22m=2e q4πϵ0 r ----(1)Now, (2p)22m=2eq4πϵ0 r1 ----(II)Equation (I)÷(ii)14=r1rr1=r4

The distance of closest approach of an $α$ - particle fired towards a nucleus with momentum $p$ ,is $r$ . The distance of closest approach when the momentum of $α$ -particle is $2 p$ is
$2 r$
$r / 4$
$4 r$
$r / 2$

$\frac{1}{2} m v^{2} = \frac{q_{1} q_{2}}{4 π ϵ_{0 r}}$

Initially, $\frac{p^{2}}{2 m} = \frac{2 e q}{4 π ϵ_{0} r}$ ----(1)

Now, $\frac{(2 p)^{2}}{2 m} = \frac{2 e q}{4 π ϵ_{0} r^{1}}$ ----(II)

Equation $(I) \div (i i)$

$\frac{1}{4} = \frac{r^{1}}{r}$

$r^{1} = \frac{r}{4}$