An alpha particle is accelerated through a potential difference of $10^6 \ V$. Its kinetic energy will be $MeV$.

Give the relationship between impact parameter and scattering angle.

An alpha particle of energy $K \text{ MeV}$ is moving towards a nucleus of atomic number $Z$. The distance of closest approach of the alpha particle to the nucleus in metres is

In $1911$,the physicist Ernest Rutherford discovered that atoms have a tiny,dense nucleus by shooting positively charged particles at a very thin gold foil. $A$ key physical property which led Rutherford to use gold was that it was

Assertion $(A)$: The impact parameter for $\alpha$-particles scattered by $180^{\circ}$ is zero.
Reason $(R)$: Zero impact parameter means that the $\alpha$-particles tend to hit the centre of the nucleus.

Who discovered the spin quantum number?