= Solution
Normalize the <probability density function> on $x\ge0$:
$$
1=N\int_0^\infty e^{-\alpha x}dx=\frac N\alpha,
\qquad
\boxed{N=\alpha,\qquad \langle x\rangle=\alpha\int_0^\infty xe^{-\alpha x}dx=\frac1\alpha.}
$$
For $y=\sqrt x\ge0$, the <change of variables> is $x=y^2$, $dx=2y\,dy$. Conservation of <probability> gives the transformed <probability density function>
$$
\boxed{g(y)=2\alpha y e^{-\alpha y^2},\quad y\ge0,\qquad g(y)=0,\quad y<0.}
$$
It is a <Rayleigh distribution> with scale $\sigma=1/\sqrt{2\alpha}$. Its derivative is $g'(y)=2\alpha e^{-\alpha y^2}(1-2\alpha y^2)$, so it rises from zero, has one maximum, and then decays:
$$
\boxed{y_{\rm mode}=\frac1{\sqrt{2\alpha}}.}
$$
The <expected value> follows by putting $z=\sqrt\alpha\,y$:
$$
\langle y\rangle=2\alpha\int_0^\infty y^2e^{-\alpha y^2}dy
=\frac2{\sqrt\alpha}\int_0^\infty z^2e^{-z^2}dz
=\boxed{\frac{\sqrt\pi}{2\sqrt\alpha}.}
$$
Here $\int_0^\infty z^2e^{-z^2}dz=\sqrt\pi/4$, obtained from the <Gaussian integral> by integration by parts. In particular $\langle\sqrt x\rangle\ne\sqrt{\langle x\rangle}$, and the <mode> changes under a nonlinear variable transformation because a density includes its <Jacobian determinant>.
\Image[/past-exam-of-the-thermal-physics-course-of-the-university-of-oxford/2016-transformed-exponential-density.png]
{title=The square root of an exponential variable has a positive density mode}
{description=Original dimensionless <Rayleigh distribution> with its <mode> and <expected value> marked. The density is measured per unit transformed variable, including the change-of-variable factor.}
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