00:01
Okay, we have four subparts for this question and we have to complete the reactions provided.
00:10
So starting with part a, we have 160 w with unknown atomic number resulting in formation of hafnium with unknown mass number and atomic number along with the unknown product.
00:30
The atomic number of hafnium and our reactants are provided, which are w equals 74, and for halfnium, the atomic number is 72 respectively.
00:54
So the equation now becomes 160, w 74, which results information of halfneum.
01:04
72 unknown plus unknown material.
01:09
We know that the sum of atomic numbers, yes, we know that the sum of atomic number and mass number on both side of equation should remain the same.
01:24
So the atomic number of unknown element can be x 74 equals 72 plus x so 74 minus 72 equals 2 equals 2 and 2 atomic number represents helium we know that the mass number for helium is 4 so this will be atomic number and we already know the mass number for helium is equal to 4.
02:07
So as of now, the helium unknown particle can be written as 4 to helium.
02:19
Now this helium is alpha particle.
02:25
As we have discussed earlier, the mass number of hafnium, so mass number for hafnium will be equal to 160 which is your initial mass number minus 4 for the helium mass number equals 156 substituting all of these values into the reaction we have 160 74 leads to formation of 156 72 hf plus 4 to 4 to moving on to our second reaction, the atomic number of chlorine and argon, atomic number of chlorine is 17 and atomic number of argon is 18.
03:30
Further, the sum of atomic number and mass number on the both side of equations should be the same.
03:37
So to determine the unknown atomic number will have 17 equals 18 plus x and the value of x turn out to be negative 1.
03:53
Now we have negative 1 atomic number only in the case of a beta particle.
04:02
Hence, we can write the equation as 38, 17, chlorine leads to formation of argon with unknown mass number, atomic number 18 plus a beta particle...