363901
\(_{92}{U^{235}}\) undergoes successive disintegrations with end product of \(_{82}{P^{203}}\). The number of \(\alpha \) and \(\beta \) particles emitted are
1 \(\alpha = 8,\,\,\beta = 6\)
2 \(\alpha = 3,\,\,\beta = 3\)
3 \(\alpha = 6,\,\,\beta = 4\)
4 \(\alpha = 6,\,\,\beta = 0\)
Explanation:
\(_{92}{U^{235}}\) disintegration end product is \(_{82}{P^{203}}\) with \(\alpha \) and \(\beta \) particles emitted. \(\Delta A = 235 - 203 = 32\) \(\therefore \) 8 alpha particles are emitted. The charge (\(Z\)) should be \(92 - 16 = 76\) But as the final charge is 82, six \({\beta ^ - }\) particles are also emitted to make the final atomic number \(Z = 82\). \(\therefore \) 8 alpha particles and six \({\beta ^ - }\) are emitted
KCET - 2008
PHXII13:NUCLEI
363902
We have seen that a gamma ray dose of \(3\,Gy\) is lethal to half the people exposed to it. If the equivalent energy were absorbed as heat, what rise in body temperature would result?
1 \(300\,\mu K\)
2 \(700\,\mu K\)
3 \(455\,\mu K\)
4 \(390\,\mu K\)
Explanation:
We can relate an absorbed energy \(Q\) and the resulting temperature increase \(\Delta T\) with relation \(Q=m r(\Delta T)\). In this equation, \(m\) is the mass of the material absorbing the energy and \(r\) is the specific heat of that material. An absorbed dose of \(3\,Gy\) corresponds to an absorbed energy per unit mass of \(3\;J\;k{g^{ - 1}}\). Let us assume that \(r\) is the specific heat of human body, is the same as that of water, \(4180\;J\;k{g^{ - 1}}\;{K^{ - 1}}\). Then, we find that \(\Delta T = \frac{{Q{\rm{/}}m}}{r} = \frac{3}{{4180}}\) \( = 7.2 \times {10^{ - 4}}\;K \simeq 700\,\mu K\)
AIIMS - 2007
PHXII13:NUCLEI
363903
A radioactive nucleus emits \(4\alpha \)- particles and \(7\beta \) - particles in succession. The ratio of number of neutrons of that of protons, is [\(A = \) mass number, \(Z = \) atomic number]
1 \(\frac{{A - Z - 13}}{{Z - 2}}\)
2 \(\frac{{A - Z - 15}}{{Z - 1}}\)
3 \(\frac{{A - Z - 13}}{{Z - 1}}\)
4 \(\frac{{A - Z - 11}}{{Z - 2}}\)
Explanation:
Let us assume, a particle \(X\) having atomic number \(Z\) and mass number \(A\) when an \(\alpha \)- particle is emitted by a nucleus, then its atomic number decreases by 2 and mass number decreases by 4. So, for given case \(_Z{X^A}{\xrightarrow{{4\alpha - particle}}_{Z - 8}}{Y^{A - 16}}\) When an \(\beta \)-particle is emitted by a nucleus its atomic number increases by one and mass number remains unchanged. So, for given case \(_{Z - 8}{Y^{A - 16}}{\xrightarrow{{7\beta - particle}}_{Z - 1}}{Z^{A - 16}}\) \(\therefore \frac{{{\rm{Number}}\,{\rm{of}}\,{\rm{neutrons}}}}{{{\rm{Number}}\,{\rm{of}}\,{\rm{protons}}}} = \frac{{\left( {A - 16} \right) - \left( {Z - 1} \right)}}{{\left( {Z - 1} \right)}}\) \( = \frac{{A - Z - 15}}{{Z - 1}}\)
MHTCET - 2020
PHXII13:NUCLEI
363904
In which of the following processes, the number of protons in the nucleus increase?
1 \(\alpha \)-decay
2 \({\beta ^ - }\)-decay
3 \({\beta ^ + }\)-decay
4 \(k-\) capture
Explanation:
For \(\alpha - decay{:_x}{A^y}{ \to _{x - 2}}{B^{y - 4}} + \alpha \) For \({\beta ^ - }{:_x}{A^y}{ \to _{x + 1}}{B^y}{ + _{ - 1}}{\beta ^0}\) For \({\beta ^ + }{:_x}{A^y}{ \to _{x - 1}}{B^y}{ + _{ + 1}}{\beta ^0}\) For \(k\)-cpature, there will be no change in the number of protons. Hence, in \(\beta \) decay.
363901
\(_{92}{U^{235}}\) undergoes successive disintegrations with end product of \(_{82}{P^{203}}\). The number of \(\alpha \) and \(\beta \) particles emitted are
1 \(\alpha = 8,\,\,\beta = 6\)
2 \(\alpha = 3,\,\,\beta = 3\)
3 \(\alpha = 6,\,\,\beta = 4\)
4 \(\alpha = 6,\,\,\beta = 0\)
Explanation:
\(_{92}{U^{235}}\) disintegration end product is \(_{82}{P^{203}}\) with \(\alpha \) and \(\beta \) particles emitted. \(\Delta A = 235 - 203 = 32\) \(\therefore \) 8 alpha particles are emitted. The charge (\(Z\)) should be \(92 - 16 = 76\) But as the final charge is 82, six \({\beta ^ - }\) particles are also emitted to make the final atomic number \(Z = 82\). \(\therefore \) 8 alpha particles and six \({\beta ^ - }\) are emitted
KCET - 2008
PHXII13:NUCLEI
363902
We have seen that a gamma ray dose of \(3\,Gy\) is lethal to half the people exposed to it. If the equivalent energy were absorbed as heat, what rise in body temperature would result?
1 \(300\,\mu K\)
2 \(700\,\mu K\)
3 \(455\,\mu K\)
4 \(390\,\mu K\)
Explanation:
We can relate an absorbed energy \(Q\) and the resulting temperature increase \(\Delta T\) with relation \(Q=m r(\Delta T)\). In this equation, \(m\) is the mass of the material absorbing the energy and \(r\) is the specific heat of that material. An absorbed dose of \(3\,Gy\) corresponds to an absorbed energy per unit mass of \(3\;J\;k{g^{ - 1}}\). Let us assume that \(r\) is the specific heat of human body, is the same as that of water, \(4180\;J\;k{g^{ - 1}}\;{K^{ - 1}}\). Then, we find that \(\Delta T = \frac{{Q{\rm{/}}m}}{r} = \frac{3}{{4180}}\) \( = 7.2 \times {10^{ - 4}}\;K \simeq 700\,\mu K\)
AIIMS - 2007
PHXII13:NUCLEI
363903
A radioactive nucleus emits \(4\alpha \)- particles and \(7\beta \) - particles in succession. The ratio of number of neutrons of that of protons, is [\(A = \) mass number, \(Z = \) atomic number]
1 \(\frac{{A - Z - 13}}{{Z - 2}}\)
2 \(\frac{{A - Z - 15}}{{Z - 1}}\)
3 \(\frac{{A - Z - 13}}{{Z - 1}}\)
4 \(\frac{{A - Z - 11}}{{Z - 2}}\)
Explanation:
Let us assume, a particle \(X\) having atomic number \(Z\) and mass number \(A\) when an \(\alpha \)- particle is emitted by a nucleus, then its atomic number decreases by 2 and mass number decreases by 4. So, for given case \(_Z{X^A}{\xrightarrow{{4\alpha - particle}}_{Z - 8}}{Y^{A - 16}}\) When an \(\beta \)-particle is emitted by a nucleus its atomic number increases by one and mass number remains unchanged. So, for given case \(_{Z - 8}{Y^{A - 16}}{\xrightarrow{{7\beta - particle}}_{Z - 1}}{Z^{A - 16}}\) \(\therefore \frac{{{\rm{Number}}\,{\rm{of}}\,{\rm{neutrons}}}}{{{\rm{Number}}\,{\rm{of}}\,{\rm{protons}}}} = \frac{{\left( {A - 16} \right) - \left( {Z - 1} \right)}}{{\left( {Z - 1} \right)}}\) \( = \frac{{A - Z - 15}}{{Z - 1}}\)
MHTCET - 2020
PHXII13:NUCLEI
363904
In which of the following processes, the number of protons in the nucleus increase?
1 \(\alpha \)-decay
2 \({\beta ^ - }\)-decay
3 \({\beta ^ + }\)-decay
4 \(k-\) capture
Explanation:
For \(\alpha - decay{:_x}{A^y}{ \to _{x - 2}}{B^{y - 4}} + \alpha \) For \({\beta ^ - }{:_x}{A^y}{ \to _{x + 1}}{B^y}{ + _{ - 1}}{\beta ^0}\) For \({\beta ^ + }{:_x}{A^y}{ \to _{x - 1}}{B^y}{ + _{ + 1}}{\beta ^0}\) For \(k\)-cpature, there will be no change in the number of protons. Hence, in \(\beta \) decay.
363901
\(_{92}{U^{235}}\) undergoes successive disintegrations with end product of \(_{82}{P^{203}}\). The number of \(\alpha \) and \(\beta \) particles emitted are
1 \(\alpha = 8,\,\,\beta = 6\)
2 \(\alpha = 3,\,\,\beta = 3\)
3 \(\alpha = 6,\,\,\beta = 4\)
4 \(\alpha = 6,\,\,\beta = 0\)
Explanation:
\(_{92}{U^{235}}\) disintegration end product is \(_{82}{P^{203}}\) with \(\alpha \) and \(\beta \) particles emitted. \(\Delta A = 235 - 203 = 32\) \(\therefore \) 8 alpha particles are emitted. The charge (\(Z\)) should be \(92 - 16 = 76\) But as the final charge is 82, six \({\beta ^ - }\) particles are also emitted to make the final atomic number \(Z = 82\). \(\therefore \) 8 alpha particles and six \({\beta ^ - }\) are emitted
KCET - 2008
PHXII13:NUCLEI
363902
We have seen that a gamma ray dose of \(3\,Gy\) is lethal to half the people exposed to it. If the equivalent energy were absorbed as heat, what rise in body temperature would result?
1 \(300\,\mu K\)
2 \(700\,\mu K\)
3 \(455\,\mu K\)
4 \(390\,\mu K\)
Explanation:
We can relate an absorbed energy \(Q\) and the resulting temperature increase \(\Delta T\) with relation \(Q=m r(\Delta T)\). In this equation, \(m\) is the mass of the material absorbing the energy and \(r\) is the specific heat of that material. An absorbed dose of \(3\,Gy\) corresponds to an absorbed energy per unit mass of \(3\;J\;k{g^{ - 1}}\). Let us assume that \(r\) is the specific heat of human body, is the same as that of water, \(4180\;J\;k{g^{ - 1}}\;{K^{ - 1}}\). Then, we find that \(\Delta T = \frac{{Q{\rm{/}}m}}{r} = \frac{3}{{4180}}\) \( = 7.2 \times {10^{ - 4}}\;K \simeq 700\,\mu K\)
AIIMS - 2007
PHXII13:NUCLEI
363903
A radioactive nucleus emits \(4\alpha \)- particles and \(7\beta \) - particles in succession. The ratio of number of neutrons of that of protons, is [\(A = \) mass number, \(Z = \) atomic number]
1 \(\frac{{A - Z - 13}}{{Z - 2}}\)
2 \(\frac{{A - Z - 15}}{{Z - 1}}\)
3 \(\frac{{A - Z - 13}}{{Z - 1}}\)
4 \(\frac{{A - Z - 11}}{{Z - 2}}\)
Explanation:
Let us assume, a particle \(X\) having atomic number \(Z\) and mass number \(A\) when an \(\alpha \)- particle is emitted by a nucleus, then its atomic number decreases by 2 and mass number decreases by 4. So, for given case \(_Z{X^A}{\xrightarrow{{4\alpha - particle}}_{Z - 8}}{Y^{A - 16}}\) When an \(\beta \)-particle is emitted by a nucleus its atomic number increases by one and mass number remains unchanged. So, for given case \(_{Z - 8}{Y^{A - 16}}{\xrightarrow{{7\beta - particle}}_{Z - 1}}{Z^{A - 16}}\) \(\therefore \frac{{{\rm{Number}}\,{\rm{of}}\,{\rm{neutrons}}}}{{{\rm{Number}}\,{\rm{of}}\,{\rm{protons}}}} = \frac{{\left( {A - 16} \right) - \left( {Z - 1} \right)}}{{\left( {Z - 1} \right)}}\) \( = \frac{{A - Z - 15}}{{Z - 1}}\)
MHTCET - 2020
PHXII13:NUCLEI
363904
In which of the following processes, the number of protons in the nucleus increase?
1 \(\alpha \)-decay
2 \({\beta ^ - }\)-decay
3 \({\beta ^ + }\)-decay
4 \(k-\) capture
Explanation:
For \(\alpha - decay{:_x}{A^y}{ \to _{x - 2}}{B^{y - 4}} + \alpha \) For \({\beta ^ - }{:_x}{A^y}{ \to _{x + 1}}{B^y}{ + _{ - 1}}{\beta ^0}\) For \({\beta ^ + }{:_x}{A^y}{ \to _{x - 1}}{B^y}{ + _{ + 1}}{\beta ^0}\) For \(k\)-cpature, there will be no change in the number of protons. Hence, in \(\beta \) decay.
363901
\(_{92}{U^{235}}\) undergoes successive disintegrations with end product of \(_{82}{P^{203}}\). The number of \(\alpha \) and \(\beta \) particles emitted are
1 \(\alpha = 8,\,\,\beta = 6\)
2 \(\alpha = 3,\,\,\beta = 3\)
3 \(\alpha = 6,\,\,\beta = 4\)
4 \(\alpha = 6,\,\,\beta = 0\)
Explanation:
\(_{92}{U^{235}}\) disintegration end product is \(_{82}{P^{203}}\) with \(\alpha \) and \(\beta \) particles emitted. \(\Delta A = 235 - 203 = 32\) \(\therefore \) 8 alpha particles are emitted. The charge (\(Z\)) should be \(92 - 16 = 76\) But as the final charge is 82, six \({\beta ^ - }\) particles are also emitted to make the final atomic number \(Z = 82\). \(\therefore \) 8 alpha particles and six \({\beta ^ - }\) are emitted
KCET - 2008
PHXII13:NUCLEI
363902
We have seen that a gamma ray dose of \(3\,Gy\) is lethal to half the people exposed to it. If the equivalent energy were absorbed as heat, what rise in body temperature would result?
1 \(300\,\mu K\)
2 \(700\,\mu K\)
3 \(455\,\mu K\)
4 \(390\,\mu K\)
Explanation:
We can relate an absorbed energy \(Q\) and the resulting temperature increase \(\Delta T\) with relation \(Q=m r(\Delta T)\). In this equation, \(m\) is the mass of the material absorbing the energy and \(r\) is the specific heat of that material. An absorbed dose of \(3\,Gy\) corresponds to an absorbed energy per unit mass of \(3\;J\;k{g^{ - 1}}\). Let us assume that \(r\) is the specific heat of human body, is the same as that of water, \(4180\;J\;k{g^{ - 1}}\;{K^{ - 1}}\). Then, we find that \(\Delta T = \frac{{Q{\rm{/}}m}}{r} = \frac{3}{{4180}}\) \( = 7.2 \times {10^{ - 4}}\;K \simeq 700\,\mu K\)
AIIMS - 2007
PHXII13:NUCLEI
363903
A radioactive nucleus emits \(4\alpha \)- particles and \(7\beta \) - particles in succession. The ratio of number of neutrons of that of protons, is [\(A = \) mass number, \(Z = \) atomic number]
1 \(\frac{{A - Z - 13}}{{Z - 2}}\)
2 \(\frac{{A - Z - 15}}{{Z - 1}}\)
3 \(\frac{{A - Z - 13}}{{Z - 1}}\)
4 \(\frac{{A - Z - 11}}{{Z - 2}}\)
Explanation:
Let us assume, a particle \(X\) having atomic number \(Z\) and mass number \(A\) when an \(\alpha \)- particle is emitted by a nucleus, then its atomic number decreases by 2 and mass number decreases by 4. So, for given case \(_Z{X^A}{\xrightarrow{{4\alpha - particle}}_{Z - 8}}{Y^{A - 16}}\) When an \(\beta \)-particle is emitted by a nucleus its atomic number increases by one and mass number remains unchanged. So, for given case \(_{Z - 8}{Y^{A - 16}}{\xrightarrow{{7\beta - particle}}_{Z - 1}}{Z^{A - 16}}\) \(\therefore \frac{{{\rm{Number}}\,{\rm{of}}\,{\rm{neutrons}}}}{{{\rm{Number}}\,{\rm{of}}\,{\rm{protons}}}} = \frac{{\left( {A - 16} \right) - \left( {Z - 1} \right)}}{{\left( {Z - 1} \right)}}\) \( = \frac{{A - Z - 15}}{{Z - 1}}\)
MHTCET - 2020
PHXII13:NUCLEI
363904
In which of the following processes, the number of protons in the nucleus increase?
1 \(\alpha \)-decay
2 \({\beta ^ - }\)-decay
3 \({\beta ^ + }\)-decay
4 \(k-\) capture
Explanation:
For \(\alpha - decay{:_x}{A^y}{ \to _{x - 2}}{B^{y - 4}} + \alpha \) For \({\beta ^ - }{:_x}{A^y}{ \to _{x + 1}}{B^y}{ + _{ - 1}}{\beta ^0}\) For \({\beta ^ + }{:_x}{A^y}{ \to _{x - 1}}{B^y}{ + _{ + 1}}{\beta ^0}\) For \(k\)-cpature, there will be no change in the number of protons. Hence, in \(\beta \) decay.
