Question

a radioactive decay equation is illustrated. Which radioactive emmission is a product of decay?

Answer 1

Answer:

The emissions of the most common forms of spontaneous radioactive decay are the alpha (α) particle, the beta (β) particle, the gamma (γ) ray, and the neutrino. The alpha particle is actually the nucleus of a helium-4 atom, with two positive charges 42He. Such charged atoms are called ions.

Answer 2

Given Limit:-

$\\\bullet\quad\displaystyle \sf \lim_{x \to 2} \dfrac{x^{2} -4}{\sqrt{x+2} -\sqrt{3x-2} }$

Solution:-

By Rationalizing the denominator,

$\\\quad\longrightarrow\quad\displaystyle \sf \lim_{x \to 2} \dfrac{x^{2} -4}{\sqrt{x+2} -\sqrt{3x-2} } \times \dfrac{\sqrt{x+2} +\sqrt{3x-2}}{\sqrt{x+2} +\sqrt{3x-2}}$

$\\\quad\longrightarrow\quad\displaystyle \sf \lim_{x \to 2} \dfrac{(x+2)(x-2)[\sqrt{x+2}+\sqrt{3x-2} \:] }{(\sqrt{x+2})^{2} -(\sqrt{3x-2})^{2} }$

$\\\quad\longrightarrow\quad\displaystyle \sf \lim_{x \to 2} \dfrac{(x+2)(x-2)[\sqrt{x+2}+\sqrt{3x-2} \:] }{ x+2-3x+2 }$

$\\\quad\longrightarrow\quad\displaystyle \sf \lim_{x \to 2} \dfrac{(x+2)(x-2)[\sqrt{x+2}+\sqrt{3x-2} \:] }{ 4-2x }$

$\\\quad\longrightarrow\quad\displaystyle \sf \lim_{x \to 2} \dfrac{(x+2)(x-2)[\sqrt{x+2}+\sqrt{3x-2} \:] }{ -2(x-2) }$

$\\\quad\longrightarrow\quad\displaystyle \sf \lim_{x \to 2} \dfrac{(x+2)[\sqrt{x+2}+\sqrt{3x-2} \:] }{ -2 }$

$\\\quad\longrightarrow\quad\sf \dfrac{(2+2)[\sqrt{2+2\:}+\sqrt{3(2)-2\:} }{-2}$

$\\\quad\longrightarrow\quad\sf \dfrac{4\times4}{-2}$

$\\\quad\therefore\quad \boxed{\frak{-8} }\bigstar$

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