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
The frequency of light having a wavelength of 425nm will be 70588 × 
.
The count of times an event takes place per unit of time is known as its frequency. The word frequency would be most frequently used to describe waves in physics including chemistry, including light, sound, including radio waves. The frequency refers to the number of times during one second that a point on a wave crosses a fixed reference point.
A waveform signal that is carried in space or down a wire has a wavelength, which is the separation between two identical places in the consecutive cycles.
Given data:
wavelength = 425nm = 
Frequency can be calculated by using the formula;
Frequency = speed of light / wavelength
Frequency = 3 × 
/ 425 × 
= 7,0588 × .
Therefore, the frequency of light having a wavelength of 425nm will be 70588 × .
To know more about frequency
brainly.com/question/18651058
#SPJ4
You should go on yahoo answers and ask it! Someone will get back to you
