While studying about Hawking radiation, I read that the last evaporation phase of a black hole is a gigantic explosion. This webpage says
... The more massive the back hole, the more energy would be released. Over time, the black hole would eventually lose so much mass that it would become small and unstable ...
So, why is the stability of a black hole inverse proportional to its size?
The text on the web page is a little misleading because black holes do not suddenly become unstable when their mass falls below some lower limit.
Black holes emit Hawking radiation and the emitted power is given by:
$$ P = \frac{\hbar c^6}{15360 \pi G^2 M^2} \tag{1} $$
where $M$ is the mass of the black hole. Note that the rate a black hole emits radiation is proportional to $1/M^2$ so the smaller the mass the faster it emits radiation. And since we know from Einstein's famous equation $E = mc^2$ that mass and energy are equivalent the smaller the mass the faster the mass decreases.
So when the black hole is large it loses mass slowly, but as it loses mass and gets smaller the rate of mass loss accelerates until in the last few seconds of its life the rate of mass loss becomes enormous.
When the web site says the black hole becomes unstable I would guess it simply means that below a certain mass the rate of evaporation becomes so high that the black hole disappears in what seems to us like an instant.
I suppose there is a sense in which black holes become unstable below a certain mass. If we use the rate that radiation is emitted to work out the temperature of a black hole we get:
$$ T = \frac{\hbar c^3}{8 \pi G M k_B} \tag{2} $$
Note that this is proportional to $1/M$ so larger black holes have lower temperatures. Now if the temperature of the black hole is less than the temperature of the cosmic microwave background (about $2.7\mathrm{K}$) the black hole will absorb radiation from the CMB and grow. If the temperature of the black hole is greater that the temperature of the CMB it will emit radiation and shrink. So we can define a black hole as unstable if its mass is lower than the value give by plugging $T = 2.7\mathrm{K}$ into equation (2).
So, why is the stability of a black hole inverse proportional to its size?
This isn't what the sentence you quoted means (I do consider that sentence to be somewhat unclear and misleading). Actually, it is fairly the opposite: large black holes have smaller temperatures, and hence they lose mass more slowly. In this sense, they are "more stable".
What the sentence does mean is that eventually the black hole becomes small, at which point its temperature is really large. At this stage, the emission of energy is so fast it eventually consumes the black hole entirely (according to Hawking's original calculation, which can be challenged at such small scales).
The comment about large masses in the original quote means that large black holes have more energy, and hence when they evaporate completely they will have lost more energy in total. Nevertheless, they lose energy at far slower rates than smaller black holes.
There is a common misconception that Hawking radiation, light emitted by black holes, is particles like electrons and high energy photons. In fact, for every black hole we're aware of, its hawking radiation consists only of unmeasurably long-wavelength photons.
The temperature of a black hole is: $$ T_H=\frac{\hbar c^3}{8\pi GMk_B} $$ And a black hole loses mass over time, because it radiates like a blackbody with that temperature. This means the most common wavelength emitted is given by: $$ \lambda=\frac{(1.3)\hbar c}{k_B T_H}=\frac{(1.3) 8\pi GM}{ c^2}=(1.3) 4\pi r_s $$ Where $r_s$ is the "Schwarzschild radius", the radius of the black hole. So typically a black hole emits light of wavelengths about ten times its radius - very low energy photons. This emitted light carries away the energy, and hence the mass of the black hole, causing it to evaporate over time. The low temperature, and long radiation wavelength, of astronomical-size black holes means that they live a very long time. An equation for the lifetime of a black hole is: $$ t_{ev}=(2.1\times10^{67}\text{ years})\left(\frac{M}{M_{\text{sun}}}\right)^3 $$ So a black hole with its mass the mass of the sun lasts $2.1\times10^{67}\text{ years}$. However, that doesn't stop us from getting an explosion for a much smaller black hole mass. A black hole that's going to evaporate in the next second has a mass of $2\times10^{5}\text{ kg}$ still, and that will emit an energy of $2\times10^{5}\text{ kg}c^2$, which is an explosion equivalent to 4.9 million megatons of TNT. Certainly enough to destroy the planet if it happened on earth, but not actually that big of an explosion on astronomical scales - much smaller than the typical supernova.
When radiation moves away from a big black hole, the gravitational time dilation that the radiation is causing on the black hole decreases. So the black hole does everything faster then, including rate of photon radiation. And including the frequency of the radiated photons.
