Tag: change in nucleus due to radioactive decay

Questions Related to change in nucleus due to radioactive decay

In radioactive decay process, the emitted negatively charged $\beta$ - particles are :

  1. the electrons present inside the nucleus

  2. the electrons produced as a result of the decay of neutrons inside the nucleus

  3. the electrons produced as a result of collisions between atoms

  4. the electrons orbiting around the nucleus


Correct Option: B
Explanation:

Beta decay can involve the emission of either electrons or positrons. The electrons or positrons emitted in a $\beta$ - decay do not exist inside the nucleus. They are only created at the time of emission, just as photons are created when an atom makes a transition from higher to a lower energy state.
In negative $\beta$- decay a neutron in the nucleus is transformed into a proton, an electron and an antineutrino. Hence, in radioactive decay process, the negatively charged emitted $\beta$- particles are the electrons produced as a result of the decay of neutrons present inside the nucleus.

Which of the following statement is correct?

  1. The rest mass of a stable nucleus is less than the sum of the rest masses of ts separated nucleons.

  2. The rest mass of a stable nucleus is greater than the sum of the rest masses of its separated nucleons

  3. In nuclear fission, energy is released by fusion two nuclei of medium mass (approximately 100 amu).

  4. In nuclear fission, energy is released by fragmentation of a very low nucleus.


Correct Option: A

Find out the missing particle in the following nuclear reaction?
$^2 _1H+^{63} _{29}Cu \rightarrow ^{64} _{30}Zn+(?)$

  1. Proton

  2. Neutron

  3. Electron

  4. Positron

  5. Deuteron


Correct Option: B
Explanation:

The given reaction :    $^2 _1 H$  $+$  $^{63} _{29} Cu  \rightarrow      ^{64} _{30} Zn  $  $+$  $^A _Z X$

In a nuclear reaction, mass number and the atomic number are conserved.
$\therefore$    $2+ 63  = 64 + A$   $\implies A = 1$
Also,  $1 + 29  = 30+Z$    $\implies Z = 0$
Thus, the missing particle is $^1 _0 n$  i.e  neutron.

The number of neutrons decreases by 1 after radioactive decay. Identify the type of decay.

  1. Alpha decay

  2. ${\beta}^{-}$ decay

  3. ${\beta}^{+}$ decay

  4. Electron capture

  5. Gamma decay


Correct Option: B
Explanation:

We know that for $\beta^-$ decay, one neutron is converted into proton and electron and the electron is ejected in the reaction. For this, the number of neutrons will decrease by one unit. 

But in $\beta^+$  decay , the positron is ejected and neutron number will increase by one unit. 

Compared to the parent nucleus, the daughter nucleus of a $\beta$ decay has:

  1. the same mass number but a smaller atomic number

  2. the same mass number but a greater atomic number

  3. a greater mass number but the same atomic number

  4. a smaller mass number but the same atomic number

  5. None of the above


Correct Option: B
Explanation:

When beta decay occurs, the atomic number of nucleus will increase by one unit and mass number will remain same. Thus, here the after beta decay the daughter nucleus has same mass number as product nuclei and one unit atomic number will increase than parent nuclei.    

When carbon $-14$ undergoes beta (electron) decay, it transmutes into what?

  1. $Carbon-13$

  2. $Carbon-12$

  3. $Nitrogen-13$

  4. $Nitrogen-12$

  5. $Nitrogen-14$


Correct Option: E
Explanation:
Beta decay of Carbon-14 : $^{14} _{6}   $ $ \rightarrow$  $^{A} _ZX $ $+$  $^{0} _{-1} e$

Conservation of mass number :  $14 = A+ 0$    $\implies A = 14$
Conservation of atomic number :  $6 = Z-1$    $\implies Z =7$
Thus the daughter nucleus is: $^{14} _{7} N$  or  $Nitrogen- 14$.

Find out the product of a $Co^{60}$ atom that undergoes one beta plus decays?

  1. iron-60

  2. manganese-60

  3. copper-60

  4. copper-62

  5. iron-62


Correct Option: A
Explanation:

$\beta^+$ decay: $^{60} _{28}Co \rightarrow$ $^{A} _{Z}X$ $+$ $^0 _{+1}e$ 

In a nuclear reaction, the mass number as well as the atomic number is conserved.
$\therefore$  $60 = 0 + A$ $\implies A =60$
Also, $27 = 1 + Z$  $\implies Z = 26$
Thus, the product of the reaction is iron-60  i.e $^{60} _{26}Fe$.

In $\beta$ decay.

  1. Atomic number decreased by one

  2. Mass number decreases by one

  3. Proton number remains the same

  4. Neutron number decreases by one


Correct Option: D
Explanation:

Beta decay of $ _Z^A X$ :    $ _Z^A X \rightarrow   _{Z+1}^A Y$ $+$ $  ^o _{-1}e$

Number of protons in X: $p _x = Z $
Number of neutrons in X: $n _x = A-Z$

Number of protons in Y: $ p _y= Z+1 $
Number of neutrons in X: $ n _y= A-(Z+1) =  A-Z - 1 $

$\implies$  $n _y = n _x -1$
Thus neutron number decreases by one in beta decay.

When an atom undergoes $\beta$-decay, its atomic number

  1. Does not change

  2. Increases by $1$

  3. Decreases by $1$

  4. Increases by $2$


Correct Option: B
Explanation:
When an atom undergoes $\beta$-decay the atomic number increases by 1.

When an atom undergoes $\beta$-decay, one of the neutrons breaks into one proton and one electron. The resultant electron is then ejected out of the nucleus and this is called as the $\beta$ particle. While the resultant proton stays inside the nucleus which results in increase of atomic number by 1, whereas the atomic mass remains invariant.

Hence the correct answer is option B.

10 grams of $^{57}Co$ kept in an open container beta-decays with a half-life of $270$ days. The weight of the material inside the container after $540$ days will be very nearly.

  1. $10g$

  2. $5g$

  3. $25g$

  4. $125g$


Correct Option: A
Explanation:

$^{57}Co$ is undergoing beta decay i.e electron is being produced.But an electron has very less mass ($9.11\times10^{-31}kg$) as compared to the $Co$ atom.therefore,after $570$ days,even the atom go under large beta decay,the weight of the material in the container will be nearly $10 g$.