147653
The distance of the closest approach of an alpha particle fired at a nucleus with kinetic energy $K$ is $r_{0}$ The distance of the closest approach when the $\alpha$ particle is fired at the same nucleus with kinetic energy $2 \mathrm{~K}$ will be
1 $2 \mathrm{r}_{0}$
2 $4 \mathrm{r}_{0}$
3 $\frac{r_{0}}{4}$
4 $\frac{r_{0}}{2}$
Explanation:
D Given, $\mathrm{r}_{1}=\mathrm{r}_{0}, \mathrm{~K}_{1}=\mathrm{K}, \mathrm{K}_{2}=2 \mathrm{~K}$ The distance of the closet approach is given by $\mathrm{r}_{0}=\frac{1}{4 \pi \varepsilon_{0}} \frac{2 \mathrm{Z}_{\mathrm{e}}^{2}}{\mathrm{~K}}$ $\mathrm{r}_{0} \propto \frac{1}{\mathrm{~K}}$ $\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}=\frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}$ $\frac{\mathrm{r}_{0}}{\mathrm{r}_{2}}=\frac{2 \mathrm{~K}}{\mathrm{~K}}$ $r_{2}=\frac{r_{0}}{2}$
GUJCET 2016
NUCLEAR PHYSICS
147654
Half- life period of a radioactive substance is 5 min, then amount of substance decayed in 20 min will be
1 $25 \%$
2 $50 \%$
3 $75 \%$
4 None of the above
Explanation:
D Given, $\mathrm{t}=20 \mathrm{~min}, \mathrm{~T}=5 \mathrm{~min}$ $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\frac{20}{5}}=\left(\frac{1}{2}\right)^{4}=\frac{1}{16}$ Fraction of decayed substance - $=1-\frac{\mathrm{N}}{\mathrm{N}_{0}}$ $=1-\left(\frac{1}{2}\right)^{4}$ $=1-\frac{1}{16}$ $=\frac{15}{16}=93.75 \%$
CG PET- 2015
NUCLEAR PHYSICS
147655
The mass number of an atom is 15 and its atomic numbers is 7 . Now, this atom absorbs an $\alpha$-particle and emits a proton. What will be the mass number of changed atom?
1 16
2 18
3 17
4 15
Explanation:
B According to question we can write ${ }_{7} \mathrm{X}^{15}+{ }_{2} \mathrm{He}^{4} \rightarrow{ }_{1} \mathrm{H}^{1}+{ }_{\mathrm{Z}} \mathrm{Y}^{\mathrm{A}}$ According to the law of conservation of charge - $7+2=1+Z$ $\mathrm{Z}=8$ According to law of conservation of mass- $15+4=1+\mathrm{A}$ $\mathrm{A}=18$ Hence, the atom has mass number 18 and atomic number is 8 .
CG PET- 2014
NUCLEAR PHYSICS
147657
In the disintegration series ${ }_{92}^{238} \mathrm{U} \stackrel{\alpha}{\longrightarrow} \mathrm{X} \stackrel{\beta}{\longrightarrow} \mathrm{Z} Y$ the values of $Z$ and $A$ respectively will be
1 92,236
2 88,230
3 90,234
4 91,234
Explanation:
D As we know that, $\alpha$-decreases mass number by 4 and reduces charge number by 2 . $A=238-4$ $A=234$ $\beta$-decays keeps mass number unchanged and increases charge by 1 . $Z=92-2+1$ $Z=91$ So, the values of $Z$ and $A$ respectively will be 91, 234.
NEET Test Series from KOTA - 10 Papers In MS WORD
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NUCLEAR PHYSICS
147653
The distance of the closest approach of an alpha particle fired at a nucleus with kinetic energy $K$ is $r_{0}$ The distance of the closest approach when the $\alpha$ particle is fired at the same nucleus with kinetic energy $2 \mathrm{~K}$ will be
1 $2 \mathrm{r}_{0}$
2 $4 \mathrm{r}_{0}$
3 $\frac{r_{0}}{4}$
4 $\frac{r_{0}}{2}$
Explanation:
D Given, $\mathrm{r}_{1}=\mathrm{r}_{0}, \mathrm{~K}_{1}=\mathrm{K}, \mathrm{K}_{2}=2 \mathrm{~K}$ The distance of the closet approach is given by $\mathrm{r}_{0}=\frac{1}{4 \pi \varepsilon_{0}} \frac{2 \mathrm{Z}_{\mathrm{e}}^{2}}{\mathrm{~K}}$ $\mathrm{r}_{0} \propto \frac{1}{\mathrm{~K}}$ $\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}=\frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}$ $\frac{\mathrm{r}_{0}}{\mathrm{r}_{2}}=\frac{2 \mathrm{~K}}{\mathrm{~K}}$ $r_{2}=\frac{r_{0}}{2}$
GUJCET 2016
NUCLEAR PHYSICS
147654
Half- life period of a radioactive substance is 5 min, then amount of substance decayed in 20 min will be
1 $25 \%$
2 $50 \%$
3 $75 \%$
4 None of the above
Explanation:
D Given, $\mathrm{t}=20 \mathrm{~min}, \mathrm{~T}=5 \mathrm{~min}$ $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\frac{20}{5}}=\left(\frac{1}{2}\right)^{4}=\frac{1}{16}$ Fraction of decayed substance - $=1-\frac{\mathrm{N}}{\mathrm{N}_{0}}$ $=1-\left(\frac{1}{2}\right)^{4}$ $=1-\frac{1}{16}$ $=\frac{15}{16}=93.75 \%$
CG PET- 2015
NUCLEAR PHYSICS
147655
The mass number of an atom is 15 and its atomic numbers is 7 . Now, this atom absorbs an $\alpha$-particle and emits a proton. What will be the mass number of changed atom?
1 16
2 18
3 17
4 15
Explanation:
B According to question we can write ${ }_{7} \mathrm{X}^{15}+{ }_{2} \mathrm{He}^{4} \rightarrow{ }_{1} \mathrm{H}^{1}+{ }_{\mathrm{Z}} \mathrm{Y}^{\mathrm{A}}$ According to the law of conservation of charge - $7+2=1+Z$ $\mathrm{Z}=8$ According to law of conservation of mass- $15+4=1+\mathrm{A}$ $\mathrm{A}=18$ Hence, the atom has mass number 18 and atomic number is 8 .
CG PET- 2014
NUCLEAR PHYSICS
147657
In the disintegration series ${ }_{92}^{238} \mathrm{U} \stackrel{\alpha}{\longrightarrow} \mathrm{X} \stackrel{\beta}{\longrightarrow} \mathrm{Z} Y$ the values of $Z$ and $A$ respectively will be
1 92,236
2 88,230
3 90,234
4 91,234
Explanation:
D As we know that, $\alpha$-decreases mass number by 4 and reduces charge number by 2 . $A=238-4$ $A=234$ $\beta$-decays keeps mass number unchanged and increases charge by 1 . $Z=92-2+1$ $Z=91$ So, the values of $Z$ and $A$ respectively will be 91, 234.
147653
The distance of the closest approach of an alpha particle fired at a nucleus with kinetic energy $K$ is $r_{0}$ The distance of the closest approach when the $\alpha$ particle is fired at the same nucleus with kinetic energy $2 \mathrm{~K}$ will be
1 $2 \mathrm{r}_{0}$
2 $4 \mathrm{r}_{0}$
3 $\frac{r_{0}}{4}$
4 $\frac{r_{0}}{2}$
Explanation:
D Given, $\mathrm{r}_{1}=\mathrm{r}_{0}, \mathrm{~K}_{1}=\mathrm{K}, \mathrm{K}_{2}=2 \mathrm{~K}$ The distance of the closet approach is given by $\mathrm{r}_{0}=\frac{1}{4 \pi \varepsilon_{0}} \frac{2 \mathrm{Z}_{\mathrm{e}}^{2}}{\mathrm{~K}}$ $\mathrm{r}_{0} \propto \frac{1}{\mathrm{~K}}$ $\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}=\frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}$ $\frac{\mathrm{r}_{0}}{\mathrm{r}_{2}}=\frac{2 \mathrm{~K}}{\mathrm{~K}}$ $r_{2}=\frac{r_{0}}{2}$
GUJCET 2016
NUCLEAR PHYSICS
147654
Half- life period of a radioactive substance is 5 min, then amount of substance decayed in 20 min will be
1 $25 \%$
2 $50 \%$
3 $75 \%$
4 None of the above
Explanation:
D Given, $\mathrm{t}=20 \mathrm{~min}, \mathrm{~T}=5 \mathrm{~min}$ $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\frac{20}{5}}=\left(\frac{1}{2}\right)^{4}=\frac{1}{16}$ Fraction of decayed substance - $=1-\frac{\mathrm{N}}{\mathrm{N}_{0}}$ $=1-\left(\frac{1}{2}\right)^{4}$ $=1-\frac{1}{16}$ $=\frac{15}{16}=93.75 \%$
CG PET- 2015
NUCLEAR PHYSICS
147655
The mass number of an atom is 15 and its atomic numbers is 7 . Now, this atom absorbs an $\alpha$-particle and emits a proton. What will be the mass number of changed atom?
1 16
2 18
3 17
4 15
Explanation:
B According to question we can write ${ }_{7} \mathrm{X}^{15}+{ }_{2} \mathrm{He}^{4} \rightarrow{ }_{1} \mathrm{H}^{1}+{ }_{\mathrm{Z}} \mathrm{Y}^{\mathrm{A}}$ According to the law of conservation of charge - $7+2=1+Z$ $\mathrm{Z}=8$ According to law of conservation of mass- $15+4=1+\mathrm{A}$ $\mathrm{A}=18$ Hence, the atom has mass number 18 and atomic number is 8 .
CG PET- 2014
NUCLEAR PHYSICS
147657
In the disintegration series ${ }_{92}^{238} \mathrm{U} \stackrel{\alpha}{\longrightarrow} \mathrm{X} \stackrel{\beta}{\longrightarrow} \mathrm{Z} Y$ the values of $Z$ and $A$ respectively will be
1 92,236
2 88,230
3 90,234
4 91,234
Explanation:
D As we know that, $\alpha$-decreases mass number by 4 and reduces charge number by 2 . $A=238-4$ $A=234$ $\beta$-decays keeps mass number unchanged and increases charge by 1 . $Z=92-2+1$ $Z=91$ So, the values of $Z$ and $A$ respectively will be 91, 234.
147653
The distance of the closest approach of an alpha particle fired at a nucleus with kinetic energy $K$ is $r_{0}$ The distance of the closest approach when the $\alpha$ particle is fired at the same nucleus with kinetic energy $2 \mathrm{~K}$ will be
1 $2 \mathrm{r}_{0}$
2 $4 \mathrm{r}_{0}$
3 $\frac{r_{0}}{4}$
4 $\frac{r_{0}}{2}$
Explanation:
D Given, $\mathrm{r}_{1}=\mathrm{r}_{0}, \mathrm{~K}_{1}=\mathrm{K}, \mathrm{K}_{2}=2 \mathrm{~K}$ The distance of the closet approach is given by $\mathrm{r}_{0}=\frac{1}{4 \pi \varepsilon_{0}} \frac{2 \mathrm{Z}_{\mathrm{e}}^{2}}{\mathrm{~K}}$ $\mathrm{r}_{0} \propto \frac{1}{\mathrm{~K}}$ $\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}=\frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}$ $\frac{\mathrm{r}_{0}}{\mathrm{r}_{2}}=\frac{2 \mathrm{~K}}{\mathrm{~K}}$ $r_{2}=\frac{r_{0}}{2}$
GUJCET 2016
NUCLEAR PHYSICS
147654
Half- life period of a radioactive substance is 5 min, then amount of substance decayed in 20 min will be
1 $25 \%$
2 $50 \%$
3 $75 \%$
4 None of the above
Explanation:
D Given, $\mathrm{t}=20 \mathrm{~min}, \mathrm{~T}=5 \mathrm{~min}$ $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\frac{20}{5}}=\left(\frac{1}{2}\right)^{4}=\frac{1}{16}$ Fraction of decayed substance - $=1-\frac{\mathrm{N}}{\mathrm{N}_{0}}$ $=1-\left(\frac{1}{2}\right)^{4}$ $=1-\frac{1}{16}$ $=\frac{15}{16}=93.75 \%$
CG PET- 2015
NUCLEAR PHYSICS
147655
The mass number of an atom is 15 and its atomic numbers is 7 . Now, this atom absorbs an $\alpha$-particle and emits a proton. What will be the mass number of changed atom?
1 16
2 18
3 17
4 15
Explanation:
B According to question we can write ${ }_{7} \mathrm{X}^{15}+{ }_{2} \mathrm{He}^{4} \rightarrow{ }_{1} \mathrm{H}^{1}+{ }_{\mathrm{Z}} \mathrm{Y}^{\mathrm{A}}$ According to the law of conservation of charge - $7+2=1+Z$ $\mathrm{Z}=8$ According to law of conservation of mass- $15+4=1+\mathrm{A}$ $\mathrm{A}=18$ Hence, the atom has mass number 18 and atomic number is 8 .
CG PET- 2014
NUCLEAR PHYSICS
147657
In the disintegration series ${ }_{92}^{238} \mathrm{U} \stackrel{\alpha}{\longrightarrow} \mathrm{X} \stackrel{\beta}{\longrightarrow} \mathrm{Z} Y$ the values of $Z$ and $A$ respectively will be
1 92,236
2 88,230
3 90,234
4 91,234
Explanation:
D As we know that, $\alpha$-decreases mass number by 4 and reduces charge number by 2 . $A=238-4$ $A=234$ $\beta$-decays keeps mass number unchanged and increases charge by 1 . $Z=92-2+1$ $Z=91$ So, the values of $Z$ and $A$ respectively will be 91, 234.