147599
In a radioactive decay, the half-life of radioactive substance is $t_{1 / 2}=69.3$ sec. The decay constant is
1 $0.01 \mathrm{sec}^{-1}$
2 $0.1 \mathrm{sec}^{-1}$
3 $10 \mathrm{sec}^{-1}$
4 $1 \mathrm{sec}^{-1}$
Explanation:
A Given, Half life $\left(\mathrm{t}_{1 / 2}\right)=69.3 \mathrm{sec}$ As we know that, $\mathrm{t}_{1 / 2}=\frac{\ln 2}{\lambda}$ $\lambda=\frac{0.693}{69.3}=0.01 \mathrm{sec}^{-1}$
COMEDK 2019
NUCLEAR PHYSICS
147606
For the following nuclear disintegration ${ }_{92}^{238} \mathrm{U} \rightarrow{ }_{82}^{206} \mathrm{~Pb}+\mathbf{x}\left[{ }_{2}^{4} \mathrm{He}\right]+6\left[{ }_{-1}^{0} \mathrm{e}\right]$ process the value of $x$ is
1 8
2 6
3 4
4 10
Explanation:
A An alpha particles has 2 neutrons and 2 protons, Thus a combined mass is 4 . So, Mass should be conserved- $238=206+4 x+0$ $4 x=32$ $x=\frac{32}{4}=8$
GUJCET 2019
NUCLEAR PHYSICS
147617
If the half-life of a radioactive elements is $\mathbf{1 0}$ $\mathrm{hr}$, its average life $=$
1 1.44
2 6.93
3 14.4
4 0.693
Explanation:
C Given, Half-life $\left(\mathrm{T}_{1 / 2}\right)=10$ hours Average life $=\frac{\text { Half life }}{\ln 2}$ $=\frac{10}{0.693}=14.4 \mathrm{hr} \text {. }$
GUJCET 2018
NUCLEAR PHYSICS
147643
In the reaction ${ }_{7} \mathrm{~N}^{14}+\alpha \rightarrow{ }_{8} \mathrm{X}^{17}+{ }_{1} \mathbf{p}^{1}$ identify $\mathrm{X}$.
1 $\mathrm{O}_{2}$
2 $\mathrm{N}_{2}$
3 $\mathrm{He}$
4 $\mathrm{Ar}$
Explanation:
A ${ }_{7} \mathrm{~N}^{14}+\alpha \rightarrow{ }_{8} \mathrm{X}^{17}+{ }_{1} \mathrm{p}^{1}$ We can write above equation as ${ }_{7} \mathrm{~N}^{14}+{ }_{2} \mathrm{He}^{4} \rightarrow{ }_{8} \mathrm{X}^{17}+{ }_{1} \mathrm{p}^{1}$ Atomic number of $\mathrm{X}$ is 8 and mass number is 17 so, oxygen $\left(\mathrm{O}_{2}\right)$ have also same atomic number and mass number.
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NUCLEAR PHYSICS
147599
In a radioactive decay, the half-life of radioactive substance is $t_{1 / 2}=69.3$ sec. The decay constant is
1 $0.01 \mathrm{sec}^{-1}$
2 $0.1 \mathrm{sec}^{-1}$
3 $10 \mathrm{sec}^{-1}$
4 $1 \mathrm{sec}^{-1}$
Explanation:
A Given, Half life $\left(\mathrm{t}_{1 / 2}\right)=69.3 \mathrm{sec}$ As we know that, $\mathrm{t}_{1 / 2}=\frac{\ln 2}{\lambda}$ $\lambda=\frac{0.693}{69.3}=0.01 \mathrm{sec}^{-1}$
COMEDK 2019
NUCLEAR PHYSICS
147606
For the following nuclear disintegration ${ }_{92}^{238} \mathrm{U} \rightarrow{ }_{82}^{206} \mathrm{~Pb}+\mathbf{x}\left[{ }_{2}^{4} \mathrm{He}\right]+6\left[{ }_{-1}^{0} \mathrm{e}\right]$ process the value of $x$ is
1 8
2 6
3 4
4 10
Explanation:
A An alpha particles has 2 neutrons and 2 protons, Thus a combined mass is 4 . So, Mass should be conserved- $238=206+4 x+0$ $4 x=32$ $x=\frac{32}{4}=8$
GUJCET 2019
NUCLEAR PHYSICS
147617
If the half-life of a radioactive elements is $\mathbf{1 0}$ $\mathrm{hr}$, its average life $=$
1 1.44
2 6.93
3 14.4
4 0.693
Explanation:
C Given, Half-life $\left(\mathrm{T}_{1 / 2}\right)=10$ hours Average life $=\frac{\text { Half life }}{\ln 2}$ $=\frac{10}{0.693}=14.4 \mathrm{hr} \text {. }$
GUJCET 2018
NUCLEAR PHYSICS
147643
In the reaction ${ }_{7} \mathrm{~N}^{14}+\alpha \rightarrow{ }_{8} \mathrm{X}^{17}+{ }_{1} \mathbf{p}^{1}$ identify $\mathrm{X}$.
1 $\mathrm{O}_{2}$
2 $\mathrm{N}_{2}$
3 $\mathrm{He}$
4 $\mathrm{Ar}$
Explanation:
A ${ }_{7} \mathrm{~N}^{14}+\alpha \rightarrow{ }_{8} \mathrm{X}^{17}+{ }_{1} \mathrm{p}^{1}$ We can write above equation as ${ }_{7} \mathrm{~N}^{14}+{ }_{2} \mathrm{He}^{4} \rightarrow{ }_{8} \mathrm{X}^{17}+{ }_{1} \mathrm{p}^{1}$ Atomic number of $\mathrm{X}$ is 8 and mass number is 17 so, oxygen $\left(\mathrm{O}_{2}\right)$ have also same atomic number and mass number.
147599
In a radioactive decay, the half-life of radioactive substance is $t_{1 / 2}=69.3$ sec. The decay constant is
1 $0.01 \mathrm{sec}^{-1}$
2 $0.1 \mathrm{sec}^{-1}$
3 $10 \mathrm{sec}^{-1}$
4 $1 \mathrm{sec}^{-1}$
Explanation:
A Given, Half life $\left(\mathrm{t}_{1 / 2}\right)=69.3 \mathrm{sec}$ As we know that, $\mathrm{t}_{1 / 2}=\frac{\ln 2}{\lambda}$ $\lambda=\frac{0.693}{69.3}=0.01 \mathrm{sec}^{-1}$
COMEDK 2019
NUCLEAR PHYSICS
147606
For the following nuclear disintegration ${ }_{92}^{238} \mathrm{U} \rightarrow{ }_{82}^{206} \mathrm{~Pb}+\mathbf{x}\left[{ }_{2}^{4} \mathrm{He}\right]+6\left[{ }_{-1}^{0} \mathrm{e}\right]$ process the value of $x$ is
1 8
2 6
3 4
4 10
Explanation:
A An alpha particles has 2 neutrons and 2 protons, Thus a combined mass is 4 . So, Mass should be conserved- $238=206+4 x+0$ $4 x=32$ $x=\frac{32}{4}=8$
GUJCET 2019
NUCLEAR PHYSICS
147617
If the half-life of a radioactive elements is $\mathbf{1 0}$ $\mathrm{hr}$, its average life $=$
1 1.44
2 6.93
3 14.4
4 0.693
Explanation:
C Given, Half-life $\left(\mathrm{T}_{1 / 2}\right)=10$ hours Average life $=\frac{\text { Half life }}{\ln 2}$ $=\frac{10}{0.693}=14.4 \mathrm{hr} \text {. }$
GUJCET 2018
NUCLEAR PHYSICS
147643
In the reaction ${ }_{7} \mathrm{~N}^{14}+\alpha \rightarrow{ }_{8} \mathrm{X}^{17}+{ }_{1} \mathbf{p}^{1}$ identify $\mathrm{X}$.
1 $\mathrm{O}_{2}$
2 $\mathrm{N}_{2}$
3 $\mathrm{He}$
4 $\mathrm{Ar}$
Explanation:
A ${ }_{7} \mathrm{~N}^{14}+\alpha \rightarrow{ }_{8} \mathrm{X}^{17}+{ }_{1} \mathrm{p}^{1}$ We can write above equation as ${ }_{7} \mathrm{~N}^{14}+{ }_{2} \mathrm{He}^{4} \rightarrow{ }_{8} \mathrm{X}^{17}+{ }_{1} \mathrm{p}^{1}$ Atomic number of $\mathrm{X}$ is 8 and mass number is 17 so, oxygen $\left(\mathrm{O}_{2}\right)$ have also same atomic number and mass number.
147599
In a radioactive decay, the half-life of radioactive substance is $t_{1 / 2}=69.3$ sec. The decay constant is
1 $0.01 \mathrm{sec}^{-1}$
2 $0.1 \mathrm{sec}^{-1}$
3 $10 \mathrm{sec}^{-1}$
4 $1 \mathrm{sec}^{-1}$
Explanation:
A Given, Half life $\left(\mathrm{t}_{1 / 2}\right)=69.3 \mathrm{sec}$ As we know that, $\mathrm{t}_{1 / 2}=\frac{\ln 2}{\lambda}$ $\lambda=\frac{0.693}{69.3}=0.01 \mathrm{sec}^{-1}$
COMEDK 2019
NUCLEAR PHYSICS
147606
For the following nuclear disintegration ${ }_{92}^{238} \mathrm{U} \rightarrow{ }_{82}^{206} \mathrm{~Pb}+\mathbf{x}\left[{ }_{2}^{4} \mathrm{He}\right]+6\left[{ }_{-1}^{0} \mathrm{e}\right]$ process the value of $x$ is
1 8
2 6
3 4
4 10
Explanation:
A An alpha particles has 2 neutrons and 2 protons, Thus a combined mass is 4 . So, Mass should be conserved- $238=206+4 x+0$ $4 x=32$ $x=\frac{32}{4}=8$
GUJCET 2019
NUCLEAR PHYSICS
147617
If the half-life of a radioactive elements is $\mathbf{1 0}$ $\mathrm{hr}$, its average life $=$
1 1.44
2 6.93
3 14.4
4 0.693
Explanation:
C Given, Half-life $\left(\mathrm{T}_{1 / 2}\right)=10$ hours Average life $=\frac{\text { Half life }}{\ln 2}$ $=\frac{10}{0.693}=14.4 \mathrm{hr} \text {. }$
GUJCET 2018
NUCLEAR PHYSICS
147643
In the reaction ${ }_{7} \mathrm{~N}^{14}+\alpha \rightarrow{ }_{8} \mathrm{X}^{17}+{ }_{1} \mathbf{p}^{1}$ identify $\mathrm{X}$.
1 $\mathrm{O}_{2}$
2 $\mathrm{N}_{2}$
3 $\mathrm{He}$
4 $\mathrm{Ar}$
Explanation:
A ${ }_{7} \mathrm{~N}^{14}+\alpha \rightarrow{ }_{8} \mathrm{X}^{17}+{ }_{1} \mathrm{p}^{1}$ We can write above equation as ${ }_{7} \mathrm{~N}^{14}+{ }_{2} \mathrm{He}^{4} \rightarrow{ }_{8} \mathrm{X}^{17}+{ }_{1} \mathrm{p}^{1}$ Atomic number of $\mathrm{X}$ is 8 and mass number is 17 so, oxygen $\left(\mathrm{O}_{2}\right)$ have also same atomic number and mass number.