${ }_{82}^{290} X \xrightarrow{\alpha} Y \xrightarrow{e^{+}} Z \xrightarrow{\beta^{-}} P \xrightarrow{e^{-}} Q$
In the nuclear emission stated above, the mass number and atomic number of the product $Q$ respectively, are
${ }_{82}^{290} X \xrightarrow{\alpha} Y \xrightarrow{e^{+}} Z \xrightarrow{\beta^{-}} P \xrightarrow{e^{-}} Q$
In the nuclear emission stated above, the mass number and atomic number of the product $Q$ respectively, are
- [NEET 2024]
- A
$286,80$
- B
$288,82$
- C
$286,81$
- D
$280,81$
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- [AIPMT 1990]
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- [AIEEE 2006]
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 tro-body dcay process, it was predicted theoretically that thekinetic 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. $n \rightarrow p+ e ^{-}+\bar{v}_{ e }$, around $1930,$ Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino $\left(\bar{v}_{ e }\right)$ to be massless and possessing negligible energy, and neutron to be at rest, momentum and energy conservation principles are applied. From this calculation, the maximum kinetic energy of the lectron 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 question $1$ and $2.$
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 tro-body dcay process, it was predicted theoretically that thekinetic 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. $n \rightarrow p+ e ^{-}+\bar{v}_{ e }$, around $1930,$ Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino $\left(\bar{v}_{ e }\right)$ to be massless and possessing negligible energy, and neutron to be at rest, momentum and energy conservation principles are applied. From this calculation, the maximum kinetic energy of the lectron 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 question $1$ and $2.$
- [IIT 2012]