In the uranium radioactive series,the initial nucleus ${}^{238}_{92}U$ decays to the final nucleus ${}^{206}_{82}Pb$. In this process,the number of $\alpha$-particles and $\beta$-particles emitted are:

  • A
    $8$ and $3$
  • B
    $16$ and $6$
  • C
    $16$ and $3$
  • D
    $8$ and $6$

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Similar Questions

$A$ nucleus with $Z = 92$ emits the following in a sequence: $\alpha, \beta^-, \beta^-, \alpha, \alpha, \alpha, \alpha, \alpha, \beta^-, \beta^-, \alpha, \beta^+, \beta^+, \alpha$. The $Z$ of the resulting nucleus is

Which of the following is in the increasing order for penetrating power?

When a neutron is disintegrated to give a $\beta$-particle,which of the following is also emitted?

The $\beta$-decay process,discovered around $1900$,is basically the decay of a neutron $(n)$. In the laboratory,a proton $(p)$ and an electron $(e^-)$ are observed as the decay products of the neutron. Therefore,considering the decay of a neutron as a two-body decay process,it was predicted theoretically that the kinetic energy of the electron should be a constant. But experimentally,it was observed that the electron kinetic energy has a continuous spectrum. Considering a three-body decay process,i.e.,$n \rightarrow p + e^- + \bar{\nu}_e$,around $1930$,Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino $(\bar{\nu}_e)$ to be massless and possessing negligible energy,and the neutron to be at rest,momentum and energy conservation principles are applied. From this calculation,the maximum kinetic energy of the electron is $0.8 \times 10^6 \ eV$. The kinetic energy carried by the proton is only the recoil energy.
$1.$ What is the maximum energy of the anti-neutrino?
$(A)$ Zero
$(B)$ Much less than $0.8 \times 10^6 \ eV$
$(C)$ Nearly $0.8 \times 10^6 \ eV$
$(D)$ Much larger than $0.8 \times 10^6 \ eV$
$2.$ If the anti-neutrino had a mass of $3 \ eV/c^2$ (where $c$ is the speed of light) instead of zero mass,what should be the range of the kinetic energy,$K$,of the electron?
$(A)$ $0 \leq K \leq 0.8 \times 10^6 \ eV$
$(B)$ $3.0 \ eV \leq K \leq 0.8 \times 10^6 \ eV$
$(C)$ $3.0 \ eV \leq K < 0.8 \times 10^6 \ eV$
$(D)$ $0 \leq K < 0.8 \times 10^6 \ eV$
Give the answer for question $1$ and $2$.

The nucleus ${ }_{Z}^{A} X$ undergoes the series of reactions given below:
${ }_{Z}^{A} X \stackrel{\alpha \text {-decay }}{\longrightarrow} P \stackrel{\beta \text {-decay }}{\longrightarrow} Q \stackrel{\alpha \text {-decay }}{\longrightarrow} R$
The number of neutrons in the nucleus $R$ is

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