$\gamma$-decay occurs when
$\gamma$-decay occurs when
- A
Pair annihilation takes place
- B
Energy is released due to conversion of neutron into proton
- C
Energy is released due to de-excitation of nucleus
- D
None of these
Similar Questions
In the given nuclear reaction, how many $\beta$ and $\alpha$ particles are emitted $_{92}{X^{235}} \to {\;_{82}}{Y^{207}}$
In the given nuclear reaction, how many $\beta$ and $\alpha$ particles are emitted $_{92}{X^{235}} \to {\;_{82}}{Y^{207}}$
Match List $I$ of the nuclear processes with List $II$ containing parent nucleus and one of the end products of each process and then select the correct answer using the codes given below the lists :
List $I$
List $II$
$P.$ $\quad$ Alpha decay
$1.$ $\quad{ }_8^{15} 0 \rightarrow{ }_7^{15} N +\ldots \ldots$.
$Q.$ $\quad$ $\beta^{+}$decay
$2.$ $\quad{ }_{92}^{238} U \rightarrow{ }_{90}^{234} Th +\ldots \ldots$.
$R.$ $\quad$Fission
$3.$ $\quad{ }_{83}^{185} Bi \rightarrow{ }_{82}^{184} Pb +\ldots \ldots$.
$S.$ $\quad$Proton emission
$4.$ $\quad{ }_{94}^{239} Pu \rightarrow{ }_{57}^{140} La +\ldots \ldots$.
Codes: $ \quad \quad P \quad Q \quad R \quad S $
Match List $I$ of the nuclear processes with List $II$ containing parent nucleus and one of the end products of each process and then select the correct answer using the codes given below the lists :
| List $I$ | List $II$ |
| $P.$ $\quad$ Alpha decay | $1.$ $\quad{ }_8^{15} 0 \rightarrow{ }_7^{15} N +\ldots \ldots$. |
| $Q.$ $\quad$ $\beta^{+}$decay | $2.$ $\quad{ }_{92}^{238} U \rightarrow{ }_{90}^{234} Th +\ldots \ldots$. |
| $R.$ $\quad$Fission | $3.$ $\quad{ }_{83}^{185} Bi \rightarrow{ }_{82}^{184} Pb +\ldots \ldots$. |
| $S.$ $\quad$Proton emission | $4.$ $\quad{ }_{94}^{239} Pu \rightarrow{ }_{57}^{140} La +\ldots \ldots$. |
Codes: $ \quad \quad P \quad Q \quad R \quad S $
- [IIT 2013]
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
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
- [AIEEE 2003]
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]
An element $A$ decays into element $C$ by a two step process :
$A \to B + {\;_2}H{e^4}$
$B \to C + \;2{e^ - }$
Then
An element $A$ decays into element $C$ by a two step process :
$A \to B + {\;_2}H{e^4}$
$B \to C + \;2{e^ - }$
Then
- [AIPMT 1989]