Answer:
\left \{ {{y=206} \atop {x=82}}Pb \right.
Explanation:
isotopes are various forms of same elements with different atomic number but different mass number.
Radioactivity is the emission of rays or particles from an atom to produce a new nuclei. There are various forms of radioactive emissions which are
- Alpha particle emission \left \{ {{y=4} \atop {x=2}}He \right.
- Beta particle emission \left \{ {{y=0} \atop {x=-1}}e \right.
- gamma radiation \left \{ {{y=0} \atop {x=0}}γ \right.
in the problem the product formed after radiation was Pb-206. isotopes of lead include Pb-204, Pb-206, Pb-207, Pb-208. they all have atomic number 82. which means the radiation cannot be ∝ or β since both radiations will alter the atomic number of the parent nucleus.
Only gamma radiation with \left \{ {{y=0} \atop {x=0}}γ \right. will produce a Pb-206 of atomic number 82 and mass number 206 , since gamma ray have 0 mass and has 0 atomic number.equation is shown below
\left \{ {{y=206} \atop {x=82}}Pb\right ⇒ \left \{ {{y=206} \atop {x=82}}Pb\right + \left \{ {{y=0} \atop {x=0}}γ\right.
Thus the atomic symbol is \left \{ {{y=206} \atop {x=82}}Pb\right
Explanation:
It is known that the maximum value of ml is equal to the vale of l. But the minimum value of n is as follows.
n = l + 1
where, n = principle quantum number
l = azimuthal quantum number
Values of n can be 1, 2, 3, 4 and so on. Whereas the values of l is 0 for s, 1 for p, 2 for d, 3 for f, and so on.
Also, "m" is known as magnetic quantum number whose values can be equal to -l and +l.
- Electronic configuration of Li is
. So here, n = 2, l = 0, m = 0 and s = ±
.
- Electronic configuration of
is
. So here, n = 4, l = 1, m = -1, 0, +1, and s = ±
.
- Electronic configuration of
is
. So here, n = 5, l = 1, m = -1, 0, +1, and s = ±
.
- Electronic configuration of B is
. So here, n = 2, l = 1, m = -1, 0, +1, and s = ±
.