I have read on Wikipedia and understood that
If $R$ is Rayleigh distributed with standard deviation = 1 then only $R^2$ has a chi-squared distribution.
Also from various sources, I had seen that if $R$ is Rayleigh distributed with standard deviation other than 1 then $R^2$ has exponential distribution.
My query is that is my understanding given in above two points correct?
Any help in this regard will be highly appreciated.
Start observing that the definition of a Reileigh $R(\sigma)$ is the following
$$R=\sqrt{X^2+Y^2}$$
where $X,Y$ are iid gaussian $N(0;\sigma^2)$
Proof:
$$f_{XY}(x,y)=\frac{1}{\sigma^22\pi}e^{-(x^2+y^2)/(2\sigma^2)}$$
passing in polars you get
$$f_{\Theta P}(\theta,\rho)=\frac{1}{2\pi}\mathbb{1}_{(0;2\pi)}(\theta)\times \frac{\rho}{\sigma^2} e^{-\rho^2/(2\sigma^2)}\mathbb{1}_{(0;\infty)}(\rho)$$
where $\Theta\sim U(0;2\pi)$ is the distribution of the angle and $P\sim\text{Rayleigh}(\sigma)$ is the distribution of the radius $R=\sqrt{X^2+Y^2}$
thus it is evident (and anyway easy to prove) that if $\sigma=1$, $R^2$ is the sum of two independent standard gaussian, that is
$$R^2(1)\sim \chi_{(2)}^2$$
It is not difficult to prove that
$$R^2(\sigma)\sim\text{Exp}\left( \frac{1}{2\sigma^2} \right)$$
Proof:
being $Y=X^2$ a monotonic transformation when $x\ge 0$ we have
$x=\sqrt{y}$; $|x'|=1/(2\sqrt{y})$ and thus
$$f_Y(y)=\frac{\sqrt{y}}{\sigma^2}\frac{1}{2\sqrt{y}}e^{-y/(2\sigma^2)}=\frac{1}{2\sigma^2}e^{-y/(2\sigma^2)}$$
Using MGF properties, immediately follows that
$$\Sigma_i R_i^2\sim\text{Gamma}\left(n; \frac{1}{2\sigma^2} \right)$$
You can find other related distribution on wiki here