147714
A nuclear reaction is given by, ${ }_{\mathrm{Z}} \mathbf{X}^{\mathrm{A}} \rightarrow{ }_{\mathrm{Z}+1} \mathbf{Y}^{\mathrm{A}}{ }_{{ }_{-1}} \mathrm{e}^{\mathbf{0}}+\bar{v}$, represents
1 fission
2 $\beta$-decay
3 $\sigma$-decay
4 fusion
Explanation:
B ${ }_{\mathrm{z}} \mathrm{X}^{\mathrm{A}} \longrightarrow \mathrm{z+1} \mathrm{Y}^{\mathrm{A}}{ }_{-1} \mathrm{e}^{0}+\bar{v}$ In the given reaction there is no change in mass number and atomic number is increased by 1 . So, it is a beta decay.
AIIMS-2012
NUCLEAR PHYSICS
147716
If Alpha, Beta and Gamma rays carry same momentum, which has the longest wavelength
1 Alpha rays
2 Beta rays
3 Gamma rays
4 None, all have same wavelength
Explanation:
D de-Broglie wavelength associated with matter is given by - $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ Where, $\lambda=$ wavelength $\mathrm{p}=\text { momentum }$ Since, alpha, beta and gamma rays carry same momentum so, these have same wavelength.
AIIMS-2006
NUCLEAR PHYSICS
147717
${ }^{238} \mathrm{U}$ has 92 protons and 238 nucleons. It decays by emitting an Alpha particle and becomes.
1 ${ }_{92}^{234} \mathrm{U}$
2 ${ }_{90}^{234} \mathrm{Th}$
3 ${ }_{92}^{235} \mathrm{U}$
4 ${ }_{93}^{237} \mathrm{~Np}$
Explanation:
B Emission of alpha particle, decrease the mass number by 4 and atomic number by 2 . $\therefore$ Decrease in mass number $=238-4=234$ Decrease in atomic number $=92-2=90$ So, ${ }^{234} \mathrm{Th}_{90}$ is emitted.
AIIMS-2006
NUCLEAR PHYSICS
147721
Two radioactive sources $X$ and $Y$ of half lives $1 \mathrm{~h}$ and $2 \mathrm{~h}$ respectively initially contain the same number of radioactive atoms. At the end of $2 \mathrm{~h}$, their rates of disintegration are in the ratio of-
1 $4: 3$
2 $3: 4$
3 $1: 2$
4 $2: 1$
Explanation:
C Rate of disintegration $\alpha$ number of atoms left So, for $\mathrm{X} \quad \frac{\mathrm{N}_{\mathrm{X}}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{2}=\frac{1}{4} \Rightarrow \mathrm{N}_{\mathrm{X}}=\frac{\mathrm{N}_{\mathrm{o}}}{4}$ for $\mathrm{Y} \quad \frac{\mathrm{N}_{\mathrm{Y}}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{1}=\frac{1}{2} \Rightarrow \mathrm{N}_{\mathrm{Y}}=\frac{\mathrm{N}_{\mathrm{o}}}{2}$ $\frac{\mathrm{N}_{\mathrm{X}}}{\mathrm{N}_{\mathrm{Y}}}=\frac{\frac{\mathrm{N}_{\mathrm{o}}}{4}}{\frac{\mathrm{N}_{\mathrm{o}}}{2}}$ $\frac{\mathrm{N}_{\mathrm{X}}}{\mathrm{N}_{\mathrm{Y}}}=\frac{1}{2}$
147714
A nuclear reaction is given by, ${ }_{\mathrm{Z}} \mathbf{X}^{\mathrm{A}} \rightarrow{ }_{\mathrm{Z}+1} \mathbf{Y}^{\mathrm{A}}{ }_{{ }_{-1}} \mathrm{e}^{\mathbf{0}}+\bar{v}$, represents
1 fission
2 $\beta$-decay
3 $\sigma$-decay
4 fusion
Explanation:
B ${ }_{\mathrm{z}} \mathrm{X}^{\mathrm{A}} \longrightarrow \mathrm{z+1} \mathrm{Y}^{\mathrm{A}}{ }_{-1} \mathrm{e}^{0}+\bar{v}$ In the given reaction there is no change in mass number and atomic number is increased by 1 . So, it is a beta decay.
AIIMS-2012
NUCLEAR PHYSICS
147716
If Alpha, Beta and Gamma rays carry same momentum, which has the longest wavelength
1 Alpha rays
2 Beta rays
3 Gamma rays
4 None, all have same wavelength
Explanation:
D de-Broglie wavelength associated with matter is given by - $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ Where, $\lambda=$ wavelength $\mathrm{p}=\text { momentum }$ Since, alpha, beta and gamma rays carry same momentum so, these have same wavelength.
AIIMS-2006
NUCLEAR PHYSICS
147717
${ }^{238} \mathrm{U}$ has 92 protons and 238 nucleons. It decays by emitting an Alpha particle and becomes.
1 ${ }_{92}^{234} \mathrm{U}$
2 ${ }_{90}^{234} \mathrm{Th}$
3 ${ }_{92}^{235} \mathrm{U}$
4 ${ }_{93}^{237} \mathrm{~Np}$
Explanation:
B Emission of alpha particle, decrease the mass number by 4 and atomic number by 2 . $\therefore$ Decrease in mass number $=238-4=234$ Decrease in atomic number $=92-2=90$ So, ${ }^{234} \mathrm{Th}_{90}$ is emitted.
AIIMS-2006
NUCLEAR PHYSICS
147721
Two radioactive sources $X$ and $Y$ of half lives $1 \mathrm{~h}$ and $2 \mathrm{~h}$ respectively initially contain the same number of radioactive atoms. At the end of $2 \mathrm{~h}$, their rates of disintegration are in the ratio of-
1 $4: 3$
2 $3: 4$
3 $1: 2$
4 $2: 1$
Explanation:
C Rate of disintegration $\alpha$ number of atoms left So, for $\mathrm{X} \quad \frac{\mathrm{N}_{\mathrm{X}}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{2}=\frac{1}{4} \Rightarrow \mathrm{N}_{\mathrm{X}}=\frac{\mathrm{N}_{\mathrm{o}}}{4}$ for $\mathrm{Y} \quad \frac{\mathrm{N}_{\mathrm{Y}}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{1}=\frac{1}{2} \Rightarrow \mathrm{N}_{\mathrm{Y}}=\frac{\mathrm{N}_{\mathrm{o}}}{2}$ $\frac{\mathrm{N}_{\mathrm{X}}}{\mathrm{N}_{\mathrm{Y}}}=\frac{\frac{\mathrm{N}_{\mathrm{o}}}{4}}{\frac{\mathrm{N}_{\mathrm{o}}}{2}}$ $\frac{\mathrm{N}_{\mathrm{X}}}{\mathrm{N}_{\mathrm{Y}}}=\frac{1}{2}$
147714
A nuclear reaction is given by, ${ }_{\mathrm{Z}} \mathbf{X}^{\mathrm{A}} \rightarrow{ }_{\mathrm{Z}+1} \mathbf{Y}^{\mathrm{A}}{ }_{{ }_{-1}} \mathrm{e}^{\mathbf{0}}+\bar{v}$, represents
1 fission
2 $\beta$-decay
3 $\sigma$-decay
4 fusion
Explanation:
B ${ }_{\mathrm{z}} \mathrm{X}^{\mathrm{A}} \longrightarrow \mathrm{z+1} \mathrm{Y}^{\mathrm{A}}{ }_{-1} \mathrm{e}^{0}+\bar{v}$ In the given reaction there is no change in mass number and atomic number is increased by 1 . So, it is a beta decay.
AIIMS-2012
NUCLEAR PHYSICS
147716
If Alpha, Beta and Gamma rays carry same momentum, which has the longest wavelength
1 Alpha rays
2 Beta rays
3 Gamma rays
4 None, all have same wavelength
Explanation:
D de-Broglie wavelength associated with matter is given by - $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ Where, $\lambda=$ wavelength $\mathrm{p}=\text { momentum }$ Since, alpha, beta and gamma rays carry same momentum so, these have same wavelength.
AIIMS-2006
NUCLEAR PHYSICS
147717
${ }^{238} \mathrm{U}$ has 92 protons and 238 nucleons. It decays by emitting an Alpha particle and becomes.
1 ${ }_{92}^{234} \mathrm{U}$
2 ${ }_{90}^{234} \mathrm{Th}$
3 ${ }_{92}^{235} \mathrm{U}$
4 ${ }_{93}^{237} \mathrm{~Np}$
Explanation:
B Emission of alpha particle, decrease the mass number by 4 and atomic number by 2 . $\therefore$ Decrease in mass number $=238-4=234$ Decrease in atomic number $=92-2=90$ So, ${ }^{234} \mathrm{Th}_{90}$ is emitted.
AIIMS-2006
NUCLEAR PHYSICS
147721
Two radioactive sources $X$ and $Y$ of half lives $1 \mathrm{~h}$ and $2 \mathrm{~h}$ respectively initially contain the same number of radioactive atoms. At the end of $2 \mathrm{~h}$, their rates of disintegration are in the ratio of-
1 $4: 3$
2 $3: 4$
3 $1: 2$
4 $2: 1$
Explanation:
C Rate of disintegration $\alpha$ number of atoms left So, for $\mathrm{X} \quad \frac{\mathrm{N}_{\mathrm{X}}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{2}=\frac{1}{4} \Rightarrow \mathrm{N}_{\mathrm{X}}=\frac{\mathrm{N}_{\mathrm{o}}}{4}$ for $\mathrm{Y} \quad \frac{\mathrm{N}_{\mathrm{Y}}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{1}=\frac{1}{2} \Rightarrow \mathrm{N}_{\mathrm{Y}}=\frac{\mathrm{N}_{\mathrm{o}}}{2}$ $\frac{\mathrm{N}_{\mathrm{X}}}{\mathrm{N}_{\mathrm{Y}}}=\frac{\frac{\mathrm{N}_{\mathrm{o}}}{4}}{\frac{\mathrm{N}_{\mathrm{o}}}{2}}$ $\frac{\mathrm{N}_{\mathrm{X}}}{\mathrm{N}_{\mathrm{Y}}}=\frac{1}{2}$
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NUCLEAR PHYSICS
147714
A nuclear reaction is given by, ${ }_{\mathrm{Z}} \mathbf{X}^{\mathrm{A}} \rightarrow{ }_{\mathrm{Z}+1} \mathbf{Y}^{\mathrm{A}}{ }_{{ }_{-1}} \mathrm{e}^{\mathbf{0}}+\bar{v}$, represents
1 fission
2 $\beta$-decay
3 $\sigma$-decay
4 fusion
Explanation:
B ${ }_{\mathrm{z}} \mathrm{X}^{\mathrm{A}} \longrightarrow \mathrm{z+1} \mathrm{Y}^{\mathrm{A}}{ }_{-1} \mathrm{e}^{0}+\bar{v}$ In the given reaction there is no change in mass number and atomic number is increased by 1 . So, it is a beta decay.
AIIMS-2012
NUCLEAR PHYSICS
147716
If Alpha, Beta and Gamma rays carry same momentum, which has the longest wavelength
1 Alpha rays
2 Beta rays
3 Gamma rays
4 None, all have same wavelength
Explanation:
D de-Broglie wavelength associated with matter is given by - $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ Where, $\lambda=$ wavelength $\mathrm{p}=\text { momentum }$ Since, alpha, beta and gamma rays carry same momentum so, these have same wavelength.
AIIMS-2006
NUCLEAR PHYSICS
147717
${ }^{238} \mathrm{U}$ has 92 protons and 238 nucleons. It decays by emitting an Alpha particle and becomes.
1 ${ }_{92}^{234} \mathrm{U}$
2 ${ }_{90}^{234} \mathrm{Th}$
3 ${ }_{92}^{235} \mathrm{U}$
4 ${ }_{93}^{237} \mathrm{~Np}$
Explanation:
B Emission of alpha particle, decrease the mass number by 4 and atomic number by 2 . $\therefore$ Decrease in mass number $=238-4=234$ Decrease in atomic number $=92-2=90$ So, ${ }^{234} \mathrm{Th}_{90}$ is emitted.
AIIMS-2006
NUCLEAR PHYSICS
147721
Two radioactive sources $X$ and $Y$ of half lives $1 \mathrm{~h}$ and $2 \mathrm{~h}$ respectively initially contain the same number of radioactive atoms. At the end of $2 \mathrm{~h}$, their rates of disintegration are in the ratio of-
1 $4: 3$
2 $3: 4$
3 $1: 2$
4 $2: 1$
Explanation:
C Rate of disintegration $\alpha$ number of atoms left So, for $\mathrm{X} \quad \frac{\mathrm{N}_{\mathrm{X}}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{2}=\frac{1}{4} \Rightarrow \mathrm{N}_{\mathrm{X}}=\frac{\mathrm{N}_{\mathrm{o}}}{4}$ for $\mathrm{Y} \quad \frac{\mathrm{N}_{\mathrm{Y}}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{1}=\frac{1}{2} \Rightarrow \mathrm{N}_{\mathrm{Y}}=\frac{\mathrm{N}_{\mathrm{o}}}{2}$ $\frac{\mathrm{N}_{\mathrm{X}}}{\mathrm{N}_{\mathrm{Y}}}=\frac{\frac{\mathrm{N}_{\mathrm{o}}}{4}}{\frac{\mathrm{N}_{\mathrm{o}}}{2}}$ $\frac{\mathrm{N}_{\mathrm{X}}}{\mathrm{N}_{\mathrm{Y}}}=\frac{1}{2}$
