Your question translates to the following:
9Be +alpha gives 12 B +?
A valid nuclear reaction must conserve atomic number and mass number, ie, if we write atomic numbers and mass numbers of elements in the reaction, the values that atomic numbers add up to on the right hand side must be same as that which they add to on the left hand side. The same of mass numbers too.
Atomic numbers : 4 + 2 = 5 + ? So, the unknown element has an atomic number of one, that's hydrogen.
Mass numbers: 9 + 4 = 12 + ? So, the unknown element(which we identified as hydrogen) has a mass number of 1, hence the missing element is 1H (a proton).
Therefore, the complete equation would be:
9 Be + 4 He gives 12 B + 1 H.
The equation for the beta decay of 17F: 917F --> 817O+ 10e + ve where the 10e is a positive beta particle or positron.
The mass of an alpha particle is approximately 4 atomic mass units (AMU), which is equivalent to 4 times the mass of a proton. In a nuclear equation, an alpha particle is represented as ( ^4_2 \text{He} ), indicating a mass number of 4 and an atomic number of 2.
224
The nuclear equation for the alpha decay of 242Pu is: ^24294Pu -> ^23892U + ^4He2 This equation shows that the nucleus of 242Pu decays into a nucleus of 238U and an alpha particle, which is a helium-4 nucleus.
If radon-210 undergoes alpha decay, it will produce the alpha particle (which is a helium-4 nucleus) and polonium-206. The equation looks like this: 86210Ra => 24He + 84206Po You'll note that in the balanced nuclear equation, the atomic numbers, which are the subscripts, balance on both sides of the equation (86 = 2 + 84). The atomic masses, which are the superscripts, also balance on both sides of the equation (210 = 4 + 206).
Neutrons are the important particles of nuclear chain reactions
Gamma rays do not have mass or charge, so they do not contribute to the balance of a nuclear equation that involves the emission of an alpha particle. The alpha particle carries away the mass and charge necessary to balance the nuclear equation.
The equation for the beta decay of 97Zr is: 4097Zr --> 4197Nb + -10e representing the beta particle as -10e.
The equation for the alpha decay of 265Bh is:107265Bh --> 105261Db + 24He where the 24He is an alpha particle or helium nucleus.
The equation for the alpha decay of 213At: 85213At --> 83209Bi + 24He where the alpha particle is represented as a helium nucleus.
The equation for the beta decay of 17F: 917F --> 817O+ 10e + ve where the 10e is a positive beta particle or positron.
The mass of an alpha particle is approximately 4 atomic mass units (AMU), which is equivalent to 4 times the mass of a proton. In a nuclear equation, an alpha particle is represented as ( ^4_2 \text{He} ), indicating a mass number of 4 and an atomic number of 2.
beta particle
To balance the nuclear equation, a beta particle (negatron) must be included. The balanced equation would be 220/88 Ra -> 4/2 He (alpha particle) + 212/86 Rn + 2 -1 e.
The equation for the beta decay of 32Si is: 1432Si --> 1532P + -10e where -10e represents a negative beta particle or electron.
The equation for the beta decay of 17F: 917F --> 817O+ 10e + ve where the 10e is a positive beta particle or positron.
The equation for the beta decay of 24Na is: 1124Na --> 1224Mg + -10e where the e is a negative beta particle or electron.