147496
The stable nucleus that has a radius half that of $\mathrm{Fe}^{56}$ is
1 $\mathrm{Li}^{7}$
2 $\mathrm{Na}^{21}$
3 $\mathrm{S}^{16}$
4 $\mathrm{Ca}^{40}$
Explanation:
A Given, $\mathrm{R}_{1}=\mathrm{R}, \mathrm{R}_{2}=\mathrm{R} / 2, \quad \mathrm{~A}_{1}=56$ We know that, $\mathrm{R}=\mathrm{R}_{0} \mathrm{~A}^{1 / 3}$ $\mathrm{R} \propto \mathrm{A}^{1 / 3}$ $\mathrm{~A} \propto \mathrm{R}^{3}$ So, $\quad \frac{A_{1}}{A_{2}}=\left(\frac{R_{1}}{R_{2}}\right)^{1 / 3}$ $\frac{56}{\mathrm{~A}_{2}}=\left(\frac{\mathrm{R}}{\mathrm{R} / 2}\right)^{3}$ $\frac{56}{\mathrm{~A}_{2}}=8$ $\mathrm{A}_{2}=\frac{56}{8}=7$ The element with mass number 7 it means $\mathrm{Li}^{7}$.
AIPMT- 1997
NUCLEAR PHYSICS
147499
What is the particle $x$ in the following nuclear reaction? ${ }_{4}^{9} \mathrm{Be}+{ }_{2}^{4} \mathrm{He} \longrightarrow{ }_{6}^{12} \mathrm{C}+\mathrm{X}$
1 Electron
2 Proton
3 Photon
4 Neutron
Explanation:
D Given the nuclear reaction ${ }_{4} \mathrm{Be}^{9}+{ }_{2} \mathrm{He}^{4} \rightarrow{ }_{6} \mathrm{C}^{12}+{ }_{\mathrm{Z}} \mathrm{X}^{\mathrm{A}}$ Sum of mass number of both side $9+4=12+A$ $A=1$ Sum of atomic number $4+2=6+Z$ $Z=0$ So, $X$ represent ${ }_{0} \mathrm{n}^{1}$ (neutron) $\text { Then, equation }{ }_{4} \mathrm{Be}^{9}+{ }_{2} \mathrm{He}^{4} \rightarrow{ }_{6} \mathrm{C}^{12}+{ }_{0} \mathrm{n}^{1}$
WB JEE 2009
NUCLEAR PHYSICS
147500
An alpha nucleus of energy $\frac{1}{2} \mathrm{mv}^{2}$ bombards a heavy nuclear target of charge $\mathrm{Ze}$. Then the distance of closest approach for the alpha nucleus will be proportional to
1 $\mathrm{v}^{2}$
2 $1 / \mathrm{m}$
3 $1 / v^{4}$
4 $1 / \mathrm{Ze}$
Explanation:
B Initial kinetic energy of the particle (K.E. $)_{\mathrm{i}}=\frac{1}{2} \mathrm{mV}^{2}$ Also initial potential energy of the system $=0$ Atomic number of alpha particle $Z_{\alpha}=2$ Final potential energy of the system at close $(\text { P.E. })_{\mathrm{f}}=\frac{\mathrm{k}(\mathrm{Ze})(2 \mathrm{e})}{\mathrm{r}}$ According to the law of conservation of energy - $(\text { K.E. })_{\mathrm{i}}+(\text { P.E. })_{\mathrm{i}}=(\text { K.E. })_{\mathrm{f}}+(\text { P.E. })_{\mathrm{f}}$ $\frac{1}{2} \mathrm{mv}^{2}+0=\frac{0+\mathrm{k} 2 \mathrm{Ze}^{2}}{\mathrm{r}}$ $\mathrm{r}=\frac{4 \mathrm{kZe}^{2}}{\mathrm{mv}^{2}}$ $\mathrm{r} \propto \frac{1}{\mathrm{~m}} \quad\left(\text { Where, } \mathrm{k}=\frac{1}{4 \pi \varepsilon_{0}}\right)$
JIPMER-2009
NUCLEAR PHYSICS
147507
Pick mirror isobars from the following: ${ }_{5} \mathrm{~B}^{12},{ }_{6} \mathrm{C}^{12},{ }_{7} \mathrm{~N}^{15},{ }_{8} \mathrm{O}^{15},{ }_{9} \mathrm{~F}^{17},{ }_{10} \mathrm{Ne}^{17}$
B Isobars of different chemical elements have different atomic numbers but have the same mass number. Mirror isobars are isobars such that the number of protons and neutrons are exchanged. ${ }_{7} \mathrm{~N}^{15},{ }_{8} \mathrm{O}^{15}$ Same mass number $=15$ Different atomic number 7, 8 ${ }_{7} \mathrm{~N}^{15}$ and ${ }_{8} \mathrm{O}^{15}$ are mirror isobars because protons and neutrons are exchanged.
147496
The stable nucleus that has a radius half that of $\mathrm{Fe}^{56}$ is
1 $\mathrm{Li}^{7}$
2 $\mathrm{Na}^{21}$
3 $\mathrm{S}^{16}$
4 $\mathrm{Ca}^{40}$
Explanation:
A Given, $\mathrm{R}_{1}=\mathrm{R}, \mathrm{R}_{2}=\mathrm{R} / 2, \quad \mathrm{~A}_{1}=56$ We know that, $\mathrm{R}=\mathrm{R}_{0} \mathrm{~A}^{1 / 3}$ $\mathrm{R} \propto \mathrm{A}^{1 / 3}$ $\mathrm{~A} \propto \mathrm{R}^{3}$ So, $\quad \frac{A_{1}}{A_{2}}=\left(\frac{R_{1}}{R_{2}}\right)^{1 / 3}$ $\frac{56}{\mathrm{~A}_{2}}=\left(\frac{\mathrm{R}}{\mathrm{R} / 2}\right)^{3}$ $\frac{56}{\mathrm{~A}_{2}}=8$ $\mathrm{A}_{2}=\frac{56}{8}=7$ The element with mass number 7 it means $\mathrm{Li}^{7}$.
AIPMT- 1997
NUCLEAR PHYSICS
147499
What is the particle $x$ in the following nuclear reaction? ${ }_{4}^{9} \mathrm{Be}+{ }_{2}^{4} \mathrm{He} \longrightarrow{ }_{6}^{12} \mathrm{C}+\mathrm{X}$
1 Electron
2 Proton
3 Photon
4 Neutron
Explanation:
D Given the nuclear reaction ${ }_{4} \mathrm{Be}^{9}+{ }_{2} \mathrm{He}^{4} \rightarrow{ }_{6} \mathrm{C}^{12}+{ }_{\mathrm{Z}} \mathrm{X}^{\mathrm{A}}$ Sum of mass number of both side $9+4=12+A$ $A=1$ Sum of atomic number $4+2=6+Z$ $Z=0$ So, $X$ represent ${ }_{0} \mathrm{n}^{1}$ (neutron) $\text { Then, equation }{ }_{4} \mathrm{Be}^{9}+{ }_{2} \mathrm{He}^{4} \rightarrow{ }_{6} \mathrm{C}^{12}+{ }_{0} \mathrm{n}^{1}$
WB JEE 2009
NUCLEAR PHYSICS
147500
An alpha nucleus of energy $\frac{1}{2} \mathrm{mv}^{2}$ bombards a heavy nuclear target of charge $\mathrm{Ze}$. Then the distance of closest approach for the alpha nucleus will be proportional to
1 $\mathrm{v}^{2}$
2 $1 / \mathrm{m}$
3 $1 / v^{4}$
4 $1 / \mathrm{Ze}$
Explanation:
B Initial kinetic energy of the particle (K.E. $)_{\mathrm{i}}=\frac{1}{2} \mathrm{mV}^{2}$ Also initial potential energy of the system $=0$ Atomic number of alpha particle $Z_{\alpha}=2$ Final potential energy of the system at close $(\text { P.E. })_{\mathrm{f}}=\frac{\mathrm{k}(\mathrm{Ze})(2 \mathrm{e})}{\mathrm{r}}$ According to the law of conservation of energy - $(\text { K.E. })_{\mathrm{i}}+(\text { P.E. })_{\mathrm{i}}=(\text { K.E. })_{\mathrm{f}}+(\text { P.E. })_{\mathrm{f}}$ $\frac{1}{2} \mathrm{mv}^{2}+0=\frac{0+\mathrm{k} 2 \mathrm{Ze}^{2}}{\mathrm{r}}$ $\mathrm{r}=\frac{4 \mathrm{kZe}^{2}}{\mathrm{mv}^{2}}$ $\mathrm{r} \propto \frac{1}{\mathrm{~m}} \quad\left(\text { Where, } \mathrm{k}=\frac{1}{4 \pi \varepsilon_{0}}\right)$
JIPMER-2009
NUCLEAR PHYSICS
147507
Pick mirror isobars from the following: ${ }_{5} \mathrm{~B}^{12},{ }_{6} \mathrm{C}^{12},{ }_{7} \mathrm{~N}^{15},{ }_{8} \mathrm{O}^{15},{ }_{9} \mathrm{~F}^{17},{ }_{10} \mathrm{Ne}^{17}$
B Isobars of different chemical elements have different atomic numbers but have the same mass number. Mirror isobars are isobars such that the number of protons and neutrons are exchanged. ${ }_{7} \mathrm{~N}^{15},{ }_{8} \mathrm{O}^{15}$ Same mass number $=15$ Different atomic number 7, 8 ${ }_{7} \mathrm{~N}^{15}$ and ${ }_{8} \mathrm{O}^{15}$ are mirror isobars because protons and neutrons are exchanged.
147496
The stable nucleus that has a radius half that of $\mathrm{Fe}^{56}$ is
1 $\mathrm{Li}^{7}$
2 $\mathrm{Na}^{21}$
3 $\mathrm{S}^{16}$
4 $\mathrm{Ca}^{40}$
Explanation:
A Given, $\mathrm{R}_{1}=\mathrm{R}, \mathrm{R}_{2}=\mathrm{R} / 2, \quad \mathrm{~A}_{1}=56$ We know that, $\mathrm{R}=\mathrm{R}_{0} \mathrm{~A}^{1 / 3}$ $\mathrm{R} \propto \mathrm{A}^{1 / 3}$ $\mathrm{~A} \propto \mathrm{R}^{3}$ So, $\quad \frac{A_{1}}{A_{2}}=\left(\frac{R_{1}}{R_{2}}\right)^{1 / 3}$ $\frac{56}{\mathrm{~A}_{2}}=\left(\frac{\mathrm{R}}{\mathrm{R} / 2}\right)^{3}$ $\frac{56}{\mathrm{~A}_{2}}=8$ $\mathrm{A}_{2}=\frac{56}{8}=7$ The element with mass number 7 it means $\mathrm{Li}^{7}$.
AIPMT- 1997
NUCLEAR PHYSICS
147499
What is the particle $x$ in the following nuclear reaction? ${ }_{4}^{9} \mathrm{Be}+{ }_{2}^{4} \mathrm{He} \longrightarrow{ }_{6}^{12} \mathrm{C}+\mathrm{X}$
1 Electron
2 Proton
3 Photon
4 Neutron
Explanation:
D Given the nuclear reaction ${ }_{4} \mathrm{Be}^{9}+{ }_{2} \mathrm{He}^{4} \rightarrow{ }_{6} \mathrm{C}^{12}+{ }_{\mathrm{Z}} \mathrm{X}^{\mathrm{A}}$ Sum of mass number of both side $9+4=12+A$ $A=1$ Sum of atomic number $4+2=6+Z$ $Z=0$ So, $X$ represent ${ }_{0} \mathrm{n}^{1}$ (neutron) $\text { Then, equation }{ }_{4} \mathrm{Be}^{9}+{ }_{2} \mathrm{He}^{4} \rightarrow{ }_{6} \mathrm{C}^{12}+{ }_{0} \mathrm{n}^{1}$
WB JEE 2009
NUCLEAR PHYSICS
147500
An alpha nucleus of energy $\frac{1}{2} \mathrm{mv}^{2}$ bombards a heavy nuclear target of charge $\mathrm{Ze}$. Then the distance of closest approach for the alpha nucleus will be proportional to
1 $\mathrm{v}^{2}$
2 $1 / \mathrm{m}$
3 $1 / v^{4}$
4 $1 / \mathrm{Ze}$
Explanation:
B Initial kinetic energy of the particle (K.E. $)_{\mathrm{i}}=\frac{1}{2} \mathrm{mV}^{2}$ Also initial potential energy of the system $=0$ Atomic number of alpha particle $Z_{\alpha}=2$ Final potential energy of the system at close $(\text { P.E. })_{\mathrm{f}}=\frac{\mathrm{k}(\mathrm{Ze})(2 \mathrm{e})}{\mathrm{r}}$ According to the law of conservation of energy - $(\text { K.E. })_{\mathrm{i}}+(\text { P.E. })_{\mathrm{i}}=(\text { K.E. })_{\mathrm{f}}+(\text { P.E. })_{\mathrm{f}}$ $\frac{1}{2} \mathrm{mv}^{2}+0=\frac{0+\mathrm{k} 2 \mathrm{Ze}^{2}}{\mathrm{r}}$ $\mathrm{r}=\frac{4 \mathrm{kZe}^{2}}{\mathrm{mv}^{2}}$ $\mathrm{r} \propto \frac{1}{\mathrm{~m}} \quad\left(\text { Where, } \mathrm{k}=\frac{1}{4 \pi \varepsilon_{0}}\right)$
JIPMER-2009
NUCLEAR PHYSICS
147507
Pick mirror isobars from the following: ${ }_{5} \mathrm{~B}^{12},{ }_{6} \mathrm{C}^{12},{ }_{7} \mathrm{~N}^{15},{ }_{8} \mathrm{O}^{15},{ }_{9} \mathrm{~F}^{17},{ }_{10} \mathrm{Ne}^{17}$
B Isobars of different chemical elements have different atomic numbers but have the same mass number. Mirror isobars are isobars such that the number of protons and neutrons are exchanged. ${ }_{7} \mathrm{~N}^{15},{ }_{8} \mathrm{O}^{15}$ Same mass number $=15$ Different atomic number 7, 8 ${ }_{7} \mathrm{~N}^{15}$ and ${ }_{8} \mathrm{O}^{15}$ are mirror isobars because protons and neutrons are exchanged.
147496
The stable nucleus that has a radius half that of $\mathrm{Fe}^{56}$ is
1 $\mathrm{Li}^{7}$
2 $\mathrm{Na}^{21}$
3 $\mathrm{S}^{16}$
4 $\mathrm{Ca}^{40}$
Explanation:
A Given, $\mathrm{R}_{1}=\mathrm{R}, \mathrm{R}_{2}=\mathrm{R} / 2, \quad \mathrm{~A}_{1}=56$ We know that, $\mathrm{R}=\mathrm{R}_{0} \mathrm{~A}^{1 / 3}$ $\mathrm{R} \propto \mathrm{A}^{1 / 3}$ $\mathrm{~A} \propto \mathrm{R}^{3}$ So, $\quad \frac{A_{1}}{A_{2}}=\left(\frac{R_{1}}{R_{2}}\right)^{1 / 3}$ $\frac{56}{\mathrm{~A}_{2}}=\left(\frac{\mathrm{R}}{\mathrm{R} / 2}\right)^{3}$ $\frac{56}{\mathrm{~A}_{2}}=8$ $\mathrm{A}_{2}=\frac{56}{8}=7$ The element with mass number 7 it means $\mathrm{Li}^{7}$.
AIPMT- 1997
NUCLEAR PHYSICS
147499
What is the particle $x$ in the following nuclear reaction? ${ }_{4}^{9} \mathrm{Be}+{ }_{2}^{4} \mathrm{He} \longrightarrow{ }_{6}^{12} \mathrm{C}+\mathrm{X}$
1 Electron
2 Proton
3 Photon
4 Neutron
Explanation:
D Given the nuclear reaction ${ }_{4} \mathrm{Be}^{9}+{ }_{2} \mathrm{He}^{4} \rightarrow{ }_{6} \mathrm{C}^{12}+{ }_{\mathrm{Z}} \mathrm{X}^{\mathrm{A}}$ Sum of mass number of both side $9+4=12+A$ $A=1$ Sum of atomic number $4+2=6+Z$ $Z=0$ So, $X$ represent ${ }_{0} \mathrm{n}^{1}$ (neutron) $\text { Then, equation }{ }_{4} \mathrm{Be}^{9}+{ }_{2} \mathrm{He}^{4} \rightarrow{ }_{6} \mathrm{C}^{12}+{ }_{0} \mathrm{n}^{1}$
WB JEE 2009
NUCLEAR PHYSICS
147500
An alpha nucleus of energy $\frac{1}{2} \mathrm{mv}^{2}$ bombards a heavy nuclear target of charge $\mathrm{Ze}$. Then the distance of closest approach for the alpha nucleus will be proportional to
1 $\mathrm{v}^{2}$
2 $1 / \mathrm{m}$
3 $1 / v^{4}$
4 $1 / \mathrm{Ze}$
Explanation:
B Initial kinetic energy of the particle (K.E. $)_{\mathrm{i}}=\frac{1}{2} \mathrm{mV}^{2}$ Also initial potential energy of the system $=0$ Atomic number of alpha particle $Z_{\alpha}=2$ Final potential energy of the system at close $(\text { P.E. })_{\mathrm{f}}=\frac{\mathrm{k}(\mathrm{Ze})(2 \mathrm{e})}{\mathrm{r}}$ According to the law of conservation of energy - $(\text { K.E. })_{\mathrm{i}}+(\text { P.E. })_{\mathrm{i}}=(\text { K.E. })_{\mathrm{f}}+(\text { P.E. })_{\mathrm{f}}$ $\frac{1}{2} \mathrm{mv}^{2}+0=\frac{0+\mathrm{k} 2 \mathrm{Ze}^{2}}{\mathrm{r}}$ $\mathrm{r}=\frac{4 \mathrm{kZe}^{2}}{\mathrm{mv}^{2}}$ $\mathrm{r} \propto \frac{1}{\mathrm{~m}} \quad\left(\text { Where, } \mathrm{k}=\frac{1}{4 \pi \varepsilon_{0}}\right)$
JIPMER-2009
NUCLEAR PHYSICS
147507
Pick mirror isobars from the following: ${ }_{5} \mathrm{~B}^{12},{ }_{6} \mathrm{C}^{12},{ }_{7} \mathrm{~N}^{15},{ }_{8} \mathrm{O}^{15},{ }_{9} \mathrm{~F}^{17},{ }_{10} \mathrm{Ne}^{17}$
B Isobars of different chemical elements have different atomic numbers but have the same mass number. Mirror isobars are isobars such that the number of protons and neutrons are exchanged. ${ }_{7} \mathrm{~N}^{15},{ }_{8} \mathrm{O}^{15}$ Same mass number $=15$ Different atomic number 7, 8 ${ }_{7} \mathrm{~N}^{15}$ and ${ }_{8} \mathrm{O}^{15}$ are mirror isobars because protons and neutrons are exchanged.
