363948
Assertion : An \(\alpha \)-particle is emitted when uranium 238 decays into thorium. Reason : The decay of uranium 238 to thorium is represented by \(_{92}^{238}U \to \,_{90}^{234}Th + \,_2^4He\). The helium nuclei is called an alpha particles.
1 Both assertion and reason are correct and reason is the correct explanation of assertion.
2 Both assertion and reason are correct but reason is not the correct explanation of assertion.
3 Assertion is correct but reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Option (1) is correct.
PHXII13:NUCLEI
363949
An atom of mass number 15 and atomic number 7 captures an \({\alpha}\)-particle and then emits a proton. The mass number and atomic number of the resulting product will respectively be
1 14 and 2
2 15 and 3
3 16 and 4
4 18 and 8
Explanation:
Given \({{ }_{7}^{15} {X}+{ }_{2}^{4} {He} \longrightarrow{ }_{Z}^{A} {Y}+{ }_{1}^{1} {H}}\) Balancing the atomic number \({(Z)}\) on both sides, \({7+2=Z+1 \Rightarrow Z=8}\) For mass number (\(A\)), \({15+4=A+1 \Rightarrow A=18}\). So correct option is (4)
PHXII13:NUCLEI
363950
\(^{238}U\) has 92 protons and 238 nucleons. If decays by emitting an alpha particle and becomes
363951
\({}_{86}{A^{222}} \to {}_{84}{B^{210}}.\) In this reaction how many \(\alpha \) and \(\beta \) particles are emitted is
1 \(\alpha ,3\beta \)
2 \(6\alpha ,3\beta \)
3 \(3\alpha ,6\beta \)
4 \(3\alpha ,4\beta \)
Explanation:
\({}_{86}^{222}A \to {}_{84}^{210}B + x{}_2^4\alpha + y{}_{ - 1}^0\beta \) By equating the atomic and mass numbers \(86 = 84 + 2x - y\) \(222 = 210 + 4x\) \( \Rightarrow x = 3,\;y = 4\)
PHXII13:NUCLEI
363952
Assertion : In \(\beta\) -decay, all the emitted electron do not have the same energy. Reason : \(\beta\) -decay is not a two body decay process.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
If it were a two-body decay process, the energy of the emitted particles would be exactly determined by momentum and energy considerations. \(\beta\) -decay involves third particle also, \(\operatorname{Viz}: \bar{\nu}(\) in \(\beta-\) decay \()\) or \(v\left(\right.\) in \(\beta^{+}-\)decay \()\). apart from \(n\) and \(p\). \(n \rightarrow p+\left(\beta^{-}\right)+\bar{v}\) Beta energy spectrum of Bismuth- 210 is shown for illustration Here \(\mathrm{r}=\) Beta-count rate \(\Rightarrow\) It is three-body process. The energy spectrum of \(\beta\)-particles is curved as shown (unlike constancy as observed for \(\alpha\)-particles). It has a maximum \(\left(E_{\max }\right)\). It indicates that most of \(\beta\)-particles exhibit \(E_{\beta} < E_{\max }\) but \(\bar{v}\) carries the remaining energy \(\left(E_{\max }-E_{\beta}\right)\). So correct option is (1).
363948
Assertion : An \(\alpha \)-particle is emitted when uranium 238 decays into thorium. Reason : The decay of uranium 238 to thorium is represented by \(_{92}^{238}U \to \,_{90}^{234}Th + \,_2^4He\). The helium nuclei is called an alpha particles.
1 Both assertion and reason are correct and reason is the correct explanation of assertion.
2 Both assertion and reason are correct but reason is not the correct explanation of assertion.
3 Assertion is correct but reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Option (1) is correct.
PHXII13:NUCLEI
363949
An atom of mass number 15 and atomic number 7 captures an \({\alpha}\)-particle and then emits a proton. The mass number and atomic number of the resulting product will respectively be
1 14 and 2
2 15 and 3
3 16 and 4
4 18 and 8
Explanation:
Given \({{ }_{7}^{15} {X}+{ }_{2}^{4} {He} \longrightarrow{ }_{Z}^{A} {Y}+{ }_{1}^{1} {H}}\) Balancing the atomic number \({(Z)}\) on both sides, \({7+2=Z+1 \Rightarrow Z=8}\) For mass number (\(A\)), \({15+4=A+1 \Rightarrow A=18}\). So correct option is (4)
PHXII13:NUCLEI
363950
\(^{238}U\) has 92 protons and 238 nucleons. If decays by emitting an alpha particle and becomes
363951
\({}_{86}{A^{222}} \to {}_{84}{B^{210}}.\) In this reaction how many \(\alpha \) and \(\beta \) particles are emitted is
1 \(\alpha ,3\beta \)
2 \(6\alpha ,3\beta \)
3 \(3\alpha ,6\beta \)
4 \(3\alpha ,4\beta \)
Explanation:
\({}_{86}^{222}A \to {}_{84}^{210}B + x{}_2^4\alpha + y{}_{ - 1}^0\beta \) By equating the atomic and mass numbers \(86 = 84 + 2x - y\) \(222 = 210 + 4x\) \( \Rightarrow x = 3,\;y = 4\)
PHXII13:NUCLEI
363952
Assertion : In \(\beta\) -decay, all the emitted electron do not have the same energy. Reason : \(\beta\) -decay is not a two body decay process.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
If it were a two-body decay process, the energy of the emitted particles would be exactly determined by momentum and energy considerations. \(\beta\) -decay involves third particle also, \(\operatorname{Viz}: \bar{\nu}(\) in \(\beta-\) decay \()\) or \(v\left(\right.\) in \(\beta^{+}-\)decay \()\). apart from \(n\) and \(p\). \(n \rightarrow p+\left(\beta^{-}\right)+\bar{v}\) Beta energy spectrum of Bismuth- 210 is shown for illustration Here \(\mathrm{r}=\) Beta-count rate \(\Rightarrow\) It is three-body process. The energy spectrum of \(\beta\)-particles is curved as shown (unlike constancy as observed for \(\alpha\)-particles). It has a maximum \(\left(E_{\max }\right)\). It indicates that most of \(\beta\)-particles exhibit \(E_{\beta} < E_{\max }\) but \(\bar{v}\) carries the remaining energy \(\left(E_{\max }-E_{\beta}\right)\). So correct option is (1).
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PHXII13:NUCLEI
363948
Assertion : An \(\alpha \)-particle is emitted when uranium 238 decays into thorium. Reason : The decay of uranium 238 to thorium is represented by \(_{92}^{238}U \to \,_{90}^{234}Th + \,_2^4He\). The helium nuclei is called an alpha particles.
1 Both assertion and reason are correct and reason is the correct explanation of assertion.
2 Both assertion and reason are correct but reason is not the correct explanation of assertion.
3 Assertion is correct but reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Option (1) is correct.
PHXII13:NUCLEI
363949
An atom of mass number 15 and atomic number 7 captures an \({\alpha}\)-particle and then emits a proton. The mass number and atomic number of the resulting product will respectively be
1 14 and 2
2 15 and 3
3 16 and 4
4 18 and 8
Explanation:
Given \({{ }_{7}^{15} {X}+{ }_{2}^{4} {He} \longrightarrow{ }_{Z}^{A} {Y}+{ }_{1}^{1} {H}}\) Balancing the atomic number \({(Z)}\) on both sides, \({7+2=Z+1 \Rightarrow Z=8}\) For mass number (\(A\)), \({15+4=A+1 \Rightarrow A=18}\). So correct option is (4)
PHXII13:NUCLEI
363950
\(^{238}U\) has 92 protons and 238 nucleons. If decays by emitting an alpha particle and becomes
363951
\({}_{86}{A^{222}} \to {}_{84}{B^{210}}.\) In this reaction how many \(\alpha \) and \(\beta \) particles are emitted is
1 \(\alpha ,3\beta \)
2 \(6\alpha ,3\beta \)
3 \(3\alpha ,6\beta \)
4 \(3\alpha ,4\beta \)
Explanation:
\({}_{86}^{222}A \to {}_{84}^{210}B + x{}_2^4\alpha + y{}_{ - 1}^0\beta \) By equating the atomic and mass numbers \(86 = 84 + 2x - y\) \(222 = 210 + 4x\) \( \Rightarrow x = 3,\;y = 4\)
PHXII13:NUCLEI
363952
Assertion : In \(\beta\) -decay, all the emitted electron do not have the same energy. Reason : \(\beta\) -decay is not a two body decay process.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
If it were a two-body decay process, the energy of the emitted particles would be exactly determined by momentum and energy considerations. \(\beta\) -decay involves third particle also, \(\operatorname{Viz}: \bar{\nu}(\) in \(\beta-\) decay \()\) or \(v\left(\right.\) in \(\beta^{+}-\)decay \()\). apart from \(n\) and \(p\). \(n \rightarrow p+\left(\beta^{-}\right)+\bar{v}\) Beta energy spectrum of Bismuth- 210 is shown for illustration Here \(\mathrm{r}=\) Beta-count rate \(\Rightarrow\) It is three-body process. The energy spectrum of \(\beta\)-particles is curved as shown (unlike constancy as observed for \(\alpha\)-particles). It has a maximum \(\left(E_{\max }\right)\). It indicates that most of \(\beta\)-particles exhibit \(E_{\beta} < E_{\max }\) but \(\bar{v}\) carries the remaining energy \(\left(E_{\max }-E_{\beta}\right)\). So correct option is (1).
363948
Assertion : An \(\alpha \)-particle is emitted when uranium 238 decays into thorium. Reason : The decay of uranium 238 to thorium is represented by \(_{92}^{238}U \to \,_{90}^{234}Th + \,_2^4He\). The helium nuclei is called an alpha particles.
1 Both assertion and reason are correct and reason is the correct explanation of assertion.
2 Both assertion and reason are correct but reason is not the correct explanation of assertion.
3 Assertion is correct but reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Option (1) is correct.
PHXII13:NUCLEI
363949
An atom of mass number 15 and atomic number 7 captures an \({\alpha}\)-particle and then emits a proton. The mass number and atomic number of the resulting product will respectively be
1 14 and 2
2 15 and 3
3 16 and 4
4 18 and 8
Explanation:
Given \({{ }_{7}^{15} {X}+{ }_{2}^{4} {He} \longrightarrow{ }_{Z}^{A} {Y}+{ }_{1}^{1} {H}}\) Balancing the atomic number \({(Z)}\) on both sides, \({7+2=Z+1 \Rightarrow Z=8}\) For mass number (\(A\)), \({15+4=A+1 \Rightarrow A=18}\). So correct option is (4)
PHXII13:NUCLEI
363950
\(^{238}U\) has 92 protons and 238 nucleons. If decays by emitting an alpha particle and becomes
363951
\({}_{86}{A^{222}} \to {}_{84}{B^{210}}.\) In this reaction how many \(\alpha \) and \(\beta \) particles are emitted is
1 \(\alpha ,3\beta \)
2 \(6\alpha ,3\beta \)
3 \(3\alpha ,6\beta \)
4 \(3\alpha ,4\beta \)
Explanation:
\({}_{86}^{222}A \to {}_{84}^{210}B + x{}_2^4\alpha + y{}_{ - 1}^0\beta \) By equating the atomic and mass numbers \(86 = 84 + 2x - y\) \(222 = 210 + 4x\) \( \Rightarrow x = 3,\;y = 4\)
PHXII13:NUCLEI
363952
Assertion : In \(\beta\) -decay, all the emitted electron do not have the same energy. Reason : \(\beta\) -decay is not a two body decay process.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
If it were a two-body decay process, the energy of the emitted particles would be exactly determined by momentum and energy considerations. \(\beta\) -decay involves third particle also, \(\operatorname{Viz}: \bar{\nu}(\) in \(\beta-\) decay \()\) or \(v\left(\right.\) in \(\beta^{+}-\)decay \()\). apart from \(n\) and \(p\). \(n \rightarrow p+\left(\beta^{-}\right)+\bar{v}\) Beta energy spectrum of Bismuth- 210 is shown for illustration Here \(\mathrm{r}=\) Beta-count rate \(\Rightarrow\) It is three-body process. The energy spectrum of \(\beta\)-particles is curved as shown (unlike constancy as observed for \(\alpha\)-particles). It has a maximum \(\left(E_{\max }\right)\). It indicates that most of \(\beta\)-particles exhibit \(E_{\beta} < E_{\max }\) but \(\bar{v}\) carries the remaining energy \(\left(E_{\max }-E_{\beta}\right)\). So correct option is (1).
363948
Assertion : An \(\alpha \)-particle is emitted when uranium 238 decays into thorium. Reason : The decay of uranium 238 to thorium is represented by \(_{92}^{238}U \to \,_{90}^{234}Th + \,_2^4He\). The helium nuclei is called an alpha particles.
1 Both assertion and reason are correct and reason is the correct explanation of assertion.
2 Both assertion and reason are correct but reason is not the correct explanation of assertion.
3 Assertion is correct but reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Option (1) is correct.
PHXII13:NUCLEI
363949
An atom of mass number 15 and atomic number 7 captures an \({\alpha}\)-particle and then emits a proton. The mass number and atomic number of the resulting product will respectively be
1 14 and 2
2 15 and 3
3 16 and 4
4 18 and 8
Explanation:
Given \({{ }_{7}^{15} {X}+{ }_{2}^{4} {He} \longrightarrow{ }_{Z}^{A} {Y}+{ }_{1}^{1} {H}}\) Balancing the atomic number \({(Z)}\) on both sides, \({7+2=Z+1 \Rightarrow Z=8}\) For mass number (\(A\)), \({15+4=A+1 \Rightarrow A=18}\). So correct option is (4)
PHXII13:NUCLEI
363950
\(^{238}U\) has 92 protons and 238 nucleons. If decays by emitting an alpha particle and becomes
363951
\({}_{86}{A^{222}} \to {}_{84}{B^{210}}.\) In this reaction how many \(\alpha \) and \(\beta \) particles are emitted is
1 \(\alpha ,3\beta \)
2 \(6\alpha ,3\beta \)
3 \(3\alpha ,6\beta \)
4 \(3\alpha ,4\beta \)
Explanation:
\({}_{86}^{222}A \to {}_{84}^{210}B + x{}_2^4\alpha + y{}_{ - 1}^0\beta \) By equating the atomic and mass numbers \(86 = 84 + 2x - y\) \(222 = 210 + 4x\) \( \Rightarrow x = 3,\;y = 4\)
PHXII13:NUCLEI
363952
Assertion : In \(\beta\) -decay, all the emitted electron do not have the same energy. Reason : \(\beta\) -decay is not a two body decay process.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
If it were a two-body decay process, the energy of the emitted particles would be exactly determined by momentum and energy considerations. \(\beta\) -decay involves third particle also, \(\operatorname{Viz}: \bar{\nu}(\) in \(\beta-\) decay \()\) or \(v\left(\right.\) in \(\beta^{+}-\)decay \()\). apart from \(n\) and \(p\). \(n \rightarrow p+\left(\beta^{-}\right)+\bar{v}\) Beta energy spectrum of Bismuth- 210 is shown for illustration Here \(\mathrm{r}=\) Beta-count rate \(\Rightarrow\) It is three-body process. The energy spectrum of \(\beta\)-particles is curved as shown (unlike constancy as observed for \(\alpha\)-particles). It has a maximum \(\left(E_{\max }\right)\). It indicates that most of \(\beta\)-particles exhibit \(E_{\beta} < E_{\max }\) but \(\bar{v}\) carries the remaining energy \(\left(E_{\max }-E_{\beta}\right)\). So correct option is (1).
