363609
In the given nuclear reaction, the element \(X \) is \(_{11}^{22}Na \to X + {e^ + } + v\)
1 \(_{10}^{23}Ne\)
2 \(_{10}^{22}Ne\)
3 \(_{12}^{22}Mg\)
4 \(_{11}^{23}Na\)
Explanation:
\(_{11}^{22}Na \to X + {e^ + } + v\) This is \({\beta ^ + } - \) decay (In this, \(A\) remains same while \(Z\) decreases by 1) \(_{11}^{22}Na \to _{10}^{22}Ne + {e^ + } + v\)
NEET - 2022
PHXII13:NUCLEI
363610
A radioactive nucleus of mass \(M\) emits a photon of frequency \(v\) and the nucleus recoils. The recoil energy will be
1 \({h^2}{v^2}/2M{c^2}\)
2 \({\rm{Zero}}\)
3 \(hv\)
4 \(M{c^2} - hv\)
Explanation:
Momentum of a photon \(p = \frac{{hv}}{c}\) Hence, recoil energy of the nucleus is \(E = \frac{{{p^2}}}{{2M}}\) (\(M\) is the mass of the nucleus) \(\therefore \quad E = \frac{{{{\left( {\frac{{hv}}{c}} \right)}^2}}}{{2M}} \Rightarrow E = \frac{{{h^2}{v^2}}}{{2M{c^2}}}\)
PHXII13:NUCLEI
363611
A nucleus at rest breaks into two nuclear parts which have their velocities ratio equal to \(2: 1\). What will be the ratio of their radii of the nuclei?
1 \(2^{1 / 3}: 1\)
2 \(1: 2^{1 / 3}\)
3 \(2^{3 / 2}: 1\)
4 \(1: 2^{3 / 2}\)
Explanation:
According to the law of conservation linear momentum, \(\begin{aligned}& m_{1} v_{1}=m_{2} v_{2} \\& \therefore \dfrac{m_{1}}{m_{2}}=\dfrac{v_{2}}{v_{1}}=\dfrac{1}{2} \quad\left(\text { given }, \dfrac{v_{1}}{v_{2}}=2\right)\end{aligned}\) Since, we know \(\begin{aligned}& R=R \cdot(A)^{1 / 3} \\& A \approx m\end{aligned}\) Thus, \(\dfrac{R_{1}}{R_{2}}=\left(\dfrac{m_{1}}{m_{2}}\right)^{1 / 3}=\left(\dfrac{1}{2}\right)^{1 / 3}\) \(\therefore R_{1}: R_{2}=1: 2^{1 / 3}\)
AIIMS - 2009
PHXII13:NUCLEI
363612
In the following reaction \(_{12}M{g^{24}} + {}_2{\rm{H}}{{\rm{e}}^4} \to {}_{14}{\rm{S}}{{\rm{i}}^X} + {}_0{n^1},X\) is
1 \(26\)
2 \(22\)
3 \(28\)
4 \(27\)
Explanation:
By equating the mass numbers. \(X + 1 = 24 + 4 \Rightarrow X = 27.\)
PHXII13:NUCLEI
363613
A stationary Thorium nucleus \((A = 200,Z = 90)\) emits an alpha particle with kinetic energy. What is the kinetic energy of the recoiling nucleus?
1 \(\frac{{{E_\alpha }}}{{108}}\)
2 \(\frac{{{E_\alpha }}}{{110}}\)
3 \(\frac{{{E_\alpha }}}{{55}}\)
4 \(\frac{{{E_\alpha }}}{{54}}\)
Explanation:
The reaction can be represented by \(_{90}^{220}Th \to _2^4He + _{88}^{216}Ra\) By conservation of momentum, \( \Rightarrow {m_\alpha }{v_\alpha } = {m_R}{v_R}\) \( \Rightarrow \frac{{{v_R}}}{{{v_\alpha }}} = \frac{{{m_\alpha }}}{{{m_R}}} = \frac{4}{{216}} = \frac{1}{{54}}\) The kinetic energy of \(Ra,\,{E_R}\), is related to the kinetic energy of alpha particle \({E_\alpha }\) by \(\frac{{{E_R}}}{{{E_\alpha }}} = \frac{{\frac{1}{2}{m_R}v_R^2}}{{\frac{1}{2}{m_\alpha }v_\alpha ^2}} = \left( {\frac{{{m_R}}}{{{m_\alpha }}}} \right){\left( {\frac{{{v_R}}}{{{v_\alpha }}}} \right)^2} = \left( {\frac{{{m_R}}}{{{m_\alpha }}}} \right){\left( {\frac{{{m_\alpha }}}{{{m_R}}}} \right)^2}\) \( = \frac{{{m_\alpha }}}{{{m_R}}} = \frac{1}{{54}}\) \(\therefore {E_R} = \frac{{{E_\alpha }}}{{54}}\)
363609
In the given nuclear reaction, the element \(X \) is \(_{11}^{22}Na \to X + {e^ + } + v\)
1 \(_{10}^{23}Ne\)
2 \(_{10}^{22}Ne\)
3 \(_{12}^{22}Mg\)
4 \(_{11}^{23}Na\)
Explanation:
\(_{11}^{22}Na \to X + {e^ + } + v\) This is \({\beta ^ + } - \) decay (In this, \(A\) remains same while \(Z\) decreases by 1) \(_{11}^{22}Na \to _{10}^{22}Ne + {e^ + } + v\)
NEET - 2022
PHXII13:NUCLEI
363610
A radioactive nucleus of mass \(M\) emits a photon of frequency \(v\) and the nucleus recoils. The recoil energy will be
1 \({h^2}{v^2}/2M{c^2}\)
2 \({\rm{Zero}}\)
3 \(hv\)
4 \(M{c^2} - hv\)
Explanation:
Momentum of a photon \(p = \frac{{hv}}{c}\) Hence, recoil energy of the nucleus is \(E = \frac{{{p^2}}}{{2M}}\) (\(M\) is the mass of the nucleus) \(\therefore \quad E = \frac{{{{\left( {\frac{{hv}}{c}} \right)}^2}}}{{2M}} \Rightarrow E = \frac{{{h^2}{v^2}}}{{2M{c^2}}}\)
PHXII13:NUCLEI
363611
A nucleus at rest breaks into two nuclear parts which have their velocities ratio equal to \(2: 1\). What will be the ratio of their radii of the nuclei?
1 \(2^{1 / 3}: 1\)
2 \(1: 2^{1 / 3}\)
3 \(2^{3 / 2}: 1\)
4 \(1: 2^{3 / 2}\)
Explanation:
According to the law of conservation linear momentum, \(\begin{aligned}& m_{1} v_{1}=m_{2} v_{2} \\& \therefore \dfrac{m_{1}}{m_{2}}=\dfrac{v_{2}}{v_{1}}=\dfrac{1}{2} \quad\left(\text { given }, \dfrac{v_{1}}{v_{2}}=2\right)\end{aligned}\) Since, we know \(\begin{aligned}& R=R \cdot(A)^{1 / 3} \\& A \approx m\end{aligned}\) Thus, \(\dfrac{R_{1}}{R_{2}}=\left(\dfrac{m_{1}}{m_{2}}\right)^{1 / 3}=\left(\dfrac{1}{2}\right)^{1 / 3}\) \(\therefore R_{1}: R_{2}=1: 2^{1 / 3}\)
AIIMS - 2009
PHXII13:NUCLEI
363612
In the following reaction \(_{12}M{g^{24}} + {}_2{\rm{H}}{{\rm{e}}^4} \to {}_{14}{\rm{S}}{{\rm{i}}^X} + {}_0{n^1},X\) is
1 \(26\)
2 \(22\)
3 \(28\)
4 \(27\)
Explanation:
By equating the mass numbers. \(X + 1 = 24 + 4 \Rightarrow X = 27.\)
PHXII13:NUCLEI
363613
A stationary Thorium nucleus \((A = 200,Z = 90)\) emits an alpha particle with kinetic energy. What is the kinetic energy of the recoiling nucleus?
1 \(\frac{{{E_\alpha }}}{{108}}\)
2 \(\frac{{{E_\alpha }}}{{110}}\)
3 \(\frac{{{E_\alpha }}}{{55}}\)
4 \(\frac{{{E_\alpha }}}{{54}}\)
Explanation:
The reaction can be represented by \(_{90}^{220}Th \to _2^4He + _{88}^{216}Ra\) By conservation of momentum, \( \Rightarrow {m_\alpha }{v_\alpha } = {m_R}{v_R}\) \( \Rightarrow \frac{{{v_R}}}{{{v_\alpha }}} = \frac{{{m_\alpha }}}{{{m_R}}} = \frac{4}{{216}} = \frac{1}{{54}}\) The kinetic energy of \(Ra,\,{E_R}\), is related to the kinetic energy of alpha particle \({E_\alpha }\) by \(\frac{{{E_R}}}{{{E_\alpha }}} = \frac{{\frac{1}{2}{m_R}v_R^2}}{{\frac{1}{2}{m_\alpha }v_\alpha ^2}} = \left( {\frac{{{m_R}}}{{{m_\alpha }}}} \right){\left( {\frac{{{v_R}}}{{{v_\alpha }}}} \right)^2} = \left( {\frac{{{m_R}}}{{{m_\alpha }}}} \right){\left( {\frac{{{m_\alpha }}}{{{m_R}}}} \right)^2}\) \( = \frac{{{m_\alpha }}}{{{m_R}}} = \frac{1}{{54}}\) \(\therefore {E_R} = \frac{{{E_\alpha }}}{{54}}\)
363609
In the given nuclear reaction, the element \(X \) is \(_{11}^{22}Na \to X + {e^ + } + v\)
1 \(_{10}^{23}Ne\)
2 \(_{10}^{22}Ne\)
3 \(_{12}^{22}Mg\)
4 \(_{11}^{23}Na\)
Explanation:
\(_{11}^{22}Na \to X + {e^ + } + v\) This is \({\beta ^ + } - \) decay (In this, \(A\) remains same while \(Z\) decreases by 1) \(_{11}^{22}Na \to _{10}^{22}Ne + {e^ + } + v\)
NEET - 2022
PHXII13:NUCLEI
363610
A radioactive nucleus of mass \(M\) emits a photon of frequency \(v\) and the nucleus recoils. The recoil energy will be
1 \({h^2}{v^2}/2M{c^2}\)
2 \({\rm{Zero}}\)
3 \(hv\)
4 \(M{c^2} - hv\)
Explanation:
Momentum of a photon \(p = \frac{{hv}}{c}\) Hence, recoil energy of the nucleus is \(E = \frac{{{p^2}}}{{2M}}\) (\(M\) is the mass of the nucleus) \(\therefore \quad E = \frac{{{{\left( {\frac{{hv}}{c}} \right)}^2}}}{{2M}} \Rightarrow E = \frac{{{h^2}{v^2}}}{{2M{c^2}}}\)
PHXII13:NUCLEI
363611
A nucleus at rest breaks into two nuclear parts which have their velocities ratio equal to \(2: 1\). What will be the ratio of their radii of the nuclei?
1 \(2^{1 / 3}: 1\)
2 \(1: 2^{1 / 3}\)
3 \(2^{3 / 2}: 1\)
4 \(1: 2^{3 / 2}\)
Explanation:
According to the law of conservation linear momentum, \(\begin{aligned}& m_{1} v_{1}=m_{2} v_{2} \\& \therefore \dfrac{m_{1}}{m_{2}}=\dfrac{v_{2}}{v_{1}}=\dfrac{1}{2} \quad\left(\text { given }, \dfrac{v_{1}}{v_{2}}=2\right)\end{aligned}\) Since, we know \(\begin{aligned}& R=R \cdot(A)^{1 / 3} \\& A \approx m\end{aligned}\) Thus, \(\dfrac{R_{1}}{R_{2}}=\left(\dfrac{m_{1}}{m_{2}}\right)^{1 / 3}=\left(\dfrac{1}{2}\right)^{1 / 3}\) \(\therefore R_{1}: R_{2}=1: 2^{1 / 3}\)
AIIMS - 2009
PHXII13:NUCLEI
363612
In the following reaction \(_{12}M{g^{24}} + {}_2{\rm{H}}{{\rm{e}}^4} \to {}_{14}{\rm{S}}{{\rm{i}}^X} + {}_0{n^1},X\) is
1 \(26\)
2 \(22\)
3 \(28\)
4 \(27\)
Explanation:
By equating the mass numbers. \(X + 1 = 24 + 4 \Rightarrow X = 27.\)
PHXII13:NUCLEI
363613
A stationary Thorium nucleus \((A = 200,Z = 90)\) emits an alpha particle with kinetic energy. What is the kinetic energy of the recoiling nucleus?
1 \(\frac{{{E_\alpha }}}{{108}}\)
2 \(\frac{{{E_\alpha }}}{{110}}\)
3 \(\frac{{{E_\alpha }}}{{55}}\)
4 \(\frac{{{E_\alpha }}}{{54}}\)
Explanation:
The reaction can be represented by \(_{90}^{220}Th \to _2^4He + _{88}^{216}Ra\) By conservation of momentum, \( \Rightarrow {m_\alpha }{v_\alpha } = {m_R}{v_R}\) \( \Rightarrow \frac{{{v_R}}}{{{v_\alpha }}} = \frac{{{m_\alpha }}}{{{m_R}}} = \frac{4}{{216}} = \frac{1}{{54}}\) The kinetic energy of \(Ra,\,{E_R}\), is related to the kinetic energy of alpha particle \({E_\alpha }\) by \(\frac{{{E_R}}}{{{E_\alpha }}} = \frac{{\frac{1}{2}{m_R}v_R^2}}{{\frac{1}{2}{m_\alpha }v_\alpha ^2}} = \left( {\frac{{{m_R}}}{{{m_\alpha }}}} \right){\left( {\frac{{{v_R}}}{{{v_\alpha }}}} \right)^2} = \left( {\frac{{{m_R}}}{{{m_\alpha }}}} \right){\left( {\frac{{{m_\alpha }}}{{{m_R}}}} \right)^2}\) \( = \frac{{{m_\alpha }}}{{{m_R}}} = \frac{1}{{54}}\) \(\therefore {E_R} = \frac{{{E_\alpha }}}{{54}}\)
363609
In the given nuclear reaction, the element \(X \) is \(_{11}^{22}Na \to X + {e^ + } + v\)
1 \(_{10}^{23}Ne\)
2 \(_{10}^{22}Ne\)
3 \(_{12}^{22}Mg\)
4 \(_{11}^{23}Na\)
Explanation:
\(_{11}^{22}Na \to X + {e^ + } + v\) This is \({\beta ^ + } - \) decay (In this, \(A\) remains same while \(Z\) decreases by 1) \(_{11}^{22}Na \to _{10}^{22}Ne + {e^ + } + v\)
NEET - 2022
PHXII13:NUCLEI
363610
A radioactive nucleus of mass \(M\) emits a photon of frequency \(v\) and the nucleus recoils. The recoil energy will be
1 \({h^2}{v^2}/2M{c^2}\)
2 \({\rm{Zero}}\)
3 \(hv\)
4 \(M{c^2} - hv\)
Explanation:
Momentum of a photon \(p = \frac{{hv}}{c}\) Hence, recoil energy of the nucleus is \(E = \frac{{{p^2}}}{{2M}}\) (\(M\) is the mass of the nucleus) \(\therefore \quad E = \frac{{{{\left( {\frac{{hv}}{c}} \right)}^2}}}{{2M}} \Rightarrow E = \frac{{{h^2}{v^2}}}{{2M{c^2}}}\)
PHXII13:NUCLEI
363611
A nucleus at rest breaks into two nuclear parts which have their velocities ratio equal to \(2: 1\). What will be the ratio of their radii of the nuclei?
1 \(2^{1 / 3}: 1\)
2 \(1: 2^{1 / 3}\)
3 \(2^{3 / 2}: 1\)
4 \(1: 2^{3 / 2}\)
Explanation:
According to the law of conservation linear momentum, \(\begin{aligned}& m_{1} v_{1}=m_{2} v_{2} \\& \therefore \dfrac{m_{1}}{m_{2}}=\dfrac{v_{2}}{v_{1}}=\dfrac{1}{2} \quad\left(\text { given }, \dfrac{v_{1}}{v_{2}}=2\right)\end{aligned}\) Since, we know \(\begin{aligned}& R=R \cdot(A)^{1 / 3} \\& A \approx m\end{aligned}\) Thus, \(\dfrac{R_{1}}{R_{2}}=\left(\dfrac{m_{1}}{m_{2}}\right)^{1 / 3}=\left(\dfrac{1}{2}\right)^{1 / 3}\) \(\therefore R_{1}: R_{2}=1: 2^{1 / 3}\)
AIIMS - 2009
PHXII13:NUCLEI
363612
In the following reaction \(_{12}M{g^{24}} + {}_2{\rm{H}}{{\rm{e}}^4} \to {}_{14}{\rm{S}}{{\rm{i}}^X} + {}_0{n^1},X\) is
1 \(26\)
2 \(22\)
3 \(28\)
4 \(27\)
Explanation:
By equating the mass numbers. \(X + 1 = 24 + 4 \Rightarrow X = 27.\)
PHXII13:NUCLEI
363613
A stationary Thorium nucleus \((A = 200,Z = 90)\) emits an alpha particle with kinetic energy. What is the kinetic energy of the recoiling nucleus?
1 \(\frac{{{E_\alpha }}}{{108}}\)
2 \(\frac{{{E_\alpha }}}{{110}}\)
3 \(\frac{{{E_\alpha }}}{{55}}\)
4 \(\frac{{{E_\alpha }}}{{54}}\)
Explanation:
The reaction can be represented by \(_{90}^{220}Th \to _2^4He + _{88}^{216}Ra\) By conservation of momentum, \( \Rightarrow {m_\alpha }{v_\alpha } = {m_R}{v_R}\) \( \Rightarrow \frac{{{v_R}}}{{{v_\alpha }}} = \frac{{{m_\alpha }}}{{{m_R}}} = \frac{4}{{216}} = \frac{1}{{54}}\) The kinetic energy of \(Ra,\,{E_R}\), is related to the kinetic energy of alpha particle \({E_\alpha }\) by \(\frac{{{E_R}}}{{{E_\alpha }}} = \frac{{\frac{1}{2}{m_R}v_R^2}}{{\frac{1}{2}{m_\alpha }v_\alpha ^2}} = \left( {\frac{{{m_R}}}{{{m_\alpha }}}} \right){\left( {\frac{{{v_R}}}{{{v_\alpha }}}} \right)^2} = \left( {\frac{{{m_R}}}{{{m_\alpha }}}} \right){\left( {\frac{{{m_\alpha }}}{{{m_R}}}} \right)^2}\) \( = \frac{{{m_\alpha }}}{{{m_R}}} = \frac{1}{{54}}\) \(\therefore {E_R} = \frac{{{E_\alpha }}}{{54}}\)
363609
In the given nuclear reaction, the element \(X \) is \(_{11}^{22}Na \to X + {e^ + } + v\)
1 \(_{10}^{23}Ne\)
2 \(_{10}^{22}Ne\)
3 \(_{12}^{22}Mg\)
4 \(_{11}^{23}Na\)
Explanation:
\(_{11}^{22}Na \to X + {e^ + } + v\) This is \({\beta ^ + } - \) decay (In this, \(A\) remains same while \(Z\) decreases by 1) \(_{11}^{22}Na \to _{10}^{22}Ne + {e^ + } + v\)
NEET - 2022
PHXII13:NUCLEI
363610
A radioactive nucleus of mass \(M\) emits a photon of frequency \(v\) and the nucleus recoils. The recoil energy will be
1 \({h^2}{v^2}/2M{c^2}\)
2 \({\rm{Zero}}\)
3 \(hv\)
4 \(M{c^2} - hv\)
Explanation:
Momentum of a photon \(p = \frac{{hv}}{c}\) Hence, recoil energy of the nucleus is \(E = \frac{{{p^2}}}{{2M}}\) (\(M\) is the mass of the nucleus) \(\therefore \quad E = \frac{{{{\left( {\frac{{hv}}{c}} \right)}^2}}}{{2M}} \Rightarrow E = \frac{{{h^2}{v^2}}}{{2M{c^2}}}\)
PHXII13:NUCLEI
363611
A nucleus at rest breaks into two nuclear parts which have their velocities ratio equal to \(2: 1\). What will be the ratio of their radii of the nuclei?
1 \(2^{1 / 3}: 1\)
2 \(1: 2^{1 / 3}\)
3 \(2^{3 / 2}: 1\)
4 \(1: 2^{3 / 2}\)
Explanation:
According to the law of conservation linear momentum, \(\begin{aligned}& m_{1} v_{1}=m_{2} v_{2} \\& \therefore \dfrac{m_{1}}{m_{2}}=\dfrac{v_{2}}{v_{1}}=\dfrac{1}{2} \quad\left(\text { given }, \dfrac{v_{1}}{v_{2}}=2\right)\end{aligned}\) Since, we know \(\begin{aligned}& R=R \cdot(A)^{1 / 3} \\& A \approx m\end{aligned}\) Thus, \(\dfrac{R_{1}}{R_{2}}=\left(\dfrac{m_{1}}{m_{2}}\right)^{1 / 3}=\left(\dfrac{1}{2}\right)^{1 / 3}\) \(\therefore R_{1}: R_{2}=1: 2^{1 / 3}\)
AIIMS - 2009
PHXII13:NUCLEI
363612
In the following reaction \(_{12}M{g^{24}} + {}_2{\rm{H}}{{\rm{e}}^4} \to {}_{14}{\rm{S}}{{\rm{i}}^X} + {}_0{n^1},X\) is
1 \(26\)
2 \(22\)
3 \(28\)
4 \(27\)
Explanation:
By equating the mass numbers. \(X + 1 = 24 + 4 \Rightarrow X = 27.\)
PHXII13:NUCLEI
363613
A stationary Thorium nucleus \((A = 200,Z = 90)\) emits an alpha particle with kinetic energy. What is the kinetic energy of the recoiling nucleus?
1 \(\frac{{{E_\alpha }}}{{108}}\)
2 \(\frac{{{E_\alpha }}}{{110}}\)
3 \(\frac{{{E_\alpha }}}{{55}}\)
4 \(\frac{{{E_\alpha }}}{{54}}\)
Explanation:
The reaction can be represented by \(_{90}^{220}Th \to _2^4He + _{88}^{216}Ra\) By conservation of momentum, \( \Rightarrow {m_\alpha }{v_\alpha } = {m_R}{v_R}\) \( \Rightarrow \frac{{{v_R}}}{{{v_\alpha }}} = \frac{{{m_\alpha }}}{{{m_R}}} = \frac{4}{{216}} = \frac{1}{{54}}\) The kinetic energy of \(Ra,\,{E_R}\), is related to the kinetic energy of alpha particle \({E_\alpha }\) by \(\frac{{{E_R}}}{{{E_\alpha }}} = \frac{{\frac{1}{2}{m_R}v_R^2}}{{\frac{1}{2}{m_\alpha }v_\alpha ^2}} = \left( {\frac{{{m_R}}}{{{m_\alpha }}}} \right){\left( {\frac{{{v_R}}}{{{v_\alpha }}}} \right)^2} = \left( {\frac{{{m_R}}}{{{m_\alpha }}}} \right){\left( {\frac{{{m_\alpha }}}{{{m_R}}}} \right)^2}\) \( = \frac{{{m_\alpha }}}{{{m_R}}} = \frac{1}{{54}}\) \(\therefore {E_R} = \frac{{{E_\alpha }}}{{54}}\)
