QBManager

NUCLEAR PHYSICS

147496 The stable nucleus that has a radius half that of ${Fe}^{56}$ is

1

{Li}^{7}

2

{Na}^{21}

3

S^{16}

4

{Ca}^{40}

Explanation:

A Given, $R_{1} = R, R_{2} = R / 2, A_{1} = 56$
We know that,
$R = R_{0} A^{1 / 3}$
$R \propto A^{1 / 3}$
$A \propto R^{3}$
So, $\frac{A_{1}}{A_{2}} = {(\frac{R_{1}}{R_{2}})}^{1 / 3}$
$\frac{56}{A_{2}} = {(\frac{R}{R / 2})}^{3}$
$\frac{56}{A_{2}} = 8$
$A_{2} = \frac{56}{8} = 7$
The element with mass number 7 it means ${Li}^{7}$ .

AIPMT- 1997

NUCLEAR PHYSICS

147499 What is the particle $x$ in the following nuclear reaction?
$_{4}^{9} Be +_{2}^{4} He ⟶_{6}^{12} C + X$

1 Electron

2 Proton

3 Photon

4 Neutron

Explanation:

D Given the nuclear reaction $_{4} {Be}^{9} +_{2} {He}^{4} \to_{6} C^{12} +_{Z} X^{A}$
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 $_{0} n^{1}$ (neutron)
$Then, equation_{4} {Be}^{9} +_{2} {He}^{4} \to_{6} C^{12} +_{0} n^{1}$

WB JEE 2009

NUCLEAR PHYSICS

147500 An alpha nucleus of energy $\frac{1}{2} {mv}^{2}$ bombards a heavy nuclear target of charge $Ze$ . Then the distance of closest approach for the alpha nucleus will be proportional to

1

v^{2}

2

1 / m

3

1 / v^{4}

4

1 / Ze

Explanation:

B Initial kinetic energy of the particle (K.E. $)_{i} = \frac{1}{2} {mV}^{2}$
Also initial potential energy of the system $= 0$
Atomic number of alpha particle $Z_{α} = 2$
Final potential energy of the system at close
$(P.E.)_{f} = \frac{k (Ze) (2 e)}{r}$
According to the law of conservation of energy -
$(K.E.)_{i} + (P.E.)_{i} = (K.E.)_{f} + (P.E.)_{f}$
$\frac{1}{2} {mv}^{2} + 0 = \frac{0 + k 2 {Ze}^{2}}{r}$
$r = \frac{4 {kZe}^{2}}{{mv}^{2}}$
$r \propto \frac{1}{m} (Where, k = \frac{1}{4 π ε_{0}})$

JIPMER-2009

NUCLEAR PHYSICS

147507 Pick mirror isobars from the following:
$_{5} B^{12},_{6} C^{12},_{7} N^{15},_{8} O^{15},_{9} F^{17},_{10} {Ne}^{17}$

1

_{5} B^{12},_{6} C^{12}

2

_{7} N^{15},_{8} O^{15}

3

_{9} F^{17},_{10} {Ne}^{17}

4

_{8} O^{15}, F^{17}

Explanation:

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.
$_{7} N^{15},_{8} O^{15}$
Same mass number $= 15$
Different atomic number 7, 8
$_{7} N^{15}$ and $_{8} O^{15}$ are mirror isobars because protons and neutrons are exchanged.

MP PET -2013

NUCLEAR PHYSICS

147496 The stable nucleus that has a radius half that of ${Fe}^{56}$ is

1

{Li}^{7}

2

{Na}^{21}

3

S^{16}

4

{Ca}^{40}

Explanation:

A Given, $R_{1} = R, R_{2} = R / 2, A_{1} = 56$
We know that,
$R = R_{0} A^{1 / 3}$
$R \propto A^{1 / 3}$
$A \propto R^{3}$
So, $\frac{A_{1}}{A_{2}} = {(\frac{R_{1}}{R_{2}})}^{1 / 3}$
$\frac{56}{A_{2}} = {(\frac{R}{R / 2})}^{3}$
$\frac{56}{A_{2}} = 8$
$A_{2} = \frac{56}{8} = 7$
The element with mass number 7 it means ${Li}^{7}$ .

AIPMT- 1997

NUCLEAR PHYSICS

147499 What is the particle $x$ in the following nuclear reaction?
$_{4}^{9} Be +_{2}^{4} He ⟶_{6}^{12} C + X$

1 Electron

2 Proton

3 Photon

4 Neutron

Explanation:

D Given the nuclear reaction $_{4} {Be}^{9} +_{2} {He}^{4} \to_{6} C^{12} +_{Z} X^{A}$
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 $_{0} n^{1}$ (neutron)
$Then, equation_{4} {Be}^{9} +_{2} {He}^{4} \to_{6} C^{12} +_{0} n^{1}$

WB JEE 2009

NUCLEAR PHYSICS

147500 An alpha nucleus of energy $\frac{1}{2} {mv}^{2}$ bombards a heavy nuclear target of charge $Ze$ . Then the distance of closest approach for the alpha nucleus will be proportional to

1

v^{2}

2

1 / m

3

1 / v^{4}

4

1 / Ze

Explanation:

B Initial kinetic energy of the particle (K.E. $)_{i} = \frac{1}{2} {mV}^{2}$
Also initial potential energy of the system $= 0$
Atomic number of alpha particle $Z_{α} = 2$
Final potential energy of the system at close
$(P.E.)_{f} = \frac{k (Ze) (2 e)}{r}$
According to the law of conservation of energy -
$(K.E.)_{i} + (P.E.)_{i} = (K.E.)_{f} + (P.E.)_{f}$
$\frac{1}{2} {mv}^{2} + 0 = \frac{0 + k 2 {Ze}^{2}}{r}$
$r = \frac{4 {kZe}^{2}}{{mv}^{2}}$
$r \propto \frac{1}{m} (Where, k = \frac{1}{4 π ε_{0}})$

JIPMER-2009

NUCLEAR PHYSICS

147507 Pick mirror isobars from the following:
$_{5} B^{12},_{6} C^{12},_{7} N^{15},_{8} O^{15},_{9} F^{17},_{10} {Ne}^{17}$

1

_{5} B^{12},_{6} C^{12}

2

_{7} N^{15},_{8} O^{15}

3

_{9} F^{17},_{10} {Ne}^{17}

4

_{8} O^{15}, F^{17}

Explanation:

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.
$_{7} N^{15},_{8} O^{15}$
Same mass number $= 15$
Different atomic number 7, 8
$_{7} N^{15}$ and $_{8} O^{15}$ are mirror isobars because protons and neutrons are exchanged.

MP PET -2013

NUCLEAR PHYSICS

147496 The stable nucleus that has a radius half that of ${Fe}^{56}$ is

1

{Li}^{7}

2

{Na}^{21}

3

S^{16}

4

{Ca}^{40}

Explanation:

A Given, $R_{1} = R, R_{2} = R / 2, A_{1} = 56$
We know that,
$R = R_{0} A^{1 / 3}$
$R \propto A^{1 / 3}$
$A \propto R^{3}$
So, $\frac{A_{1}}{A_{2}} = {(\frac{R_{1}}{R_{2}})}^{1 / 3}$
$\frac{56}{A_{2}} = {(\frac{R}{R / 2})}^{3}$
$\frac{56}{A_{2}} = 8$
$A_{2} = \frac{56}{8} = 7$
The element with mass number 7 it means ${Li}^{7}$ .

AIPMT- 1997

NUCLEAR PHYSICS

147499 What is the particle $x$ in the following nuclear reaction?
$_{4}^{9} Be +_{2}^{4} He ⟶_{6}^{12} C + X$

1 Electron

2 Proton

3 Photon

4 Neutron

Explanation:

D Given the nuclear reaction $_{4} {Be}^{9} +_{2} {He}^{4} \to_{6} C^{12} +_{Z} X^{A}$
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 $_{0} n^{1}$ (neutron)
$Then, equation_{4} {Be}^{9} +_{2} {He}^{4} \to_{6} C^{12} +_{0} n^{1}$

WB JEE 2009

NUCLEAR PHYSICS

147500 An alpha nucleus of energy $\frac{1}{2} {mv}^{2}$ bombards a heavy nuclear target of charge $Ze$ . Then the distance of closest approach for the alpha nucleus will be proportional to

1

v^{2}

2

1 / m

3

1 / v^{4}

4

1 / Ze

Explanation:

B Initial kinetic energy of the particle (K.E. $)_{i} = \frac{1}{2} {mV}^{2}$
Also initial potential energy of the system $= 0$
Atomic number of alpha particle $Z_{α} = 2$
Final potential energy of the system at close
$(P.E.)_{f} = \frac{k (Ze) (2 e)}{r}$
According to the law of conservation of energy -
$(K.E.)_{i} + (P.E.)_{i} = (K.E.)_{f} + (P.E.)_{f}$
$\frac{1}{2} {mv}^{2} + 0 = \frac{0 + k 2 {Ze}^{2}}{r}$
$r = \frac{4 {kZe}^{2}}{{mv}^{2}}$
$r \propto \frac{1}{m} (Where, k = \frac{1}{4 π ε_{0}})$

JIPMER-2009

NUCLEAR PHYSICS

147507 Pick mirror isobars from the following:
$_{5} B^{12},_{6} C^{12},_{7} N^{15},_{8} O^{15},_{9} F^{17},_{10} {Ne}^{17}$

1

_{5} B^{12},_{6} C^{12}

2

_{7} N^{15},_{8} O^{15}

3

_{9} F^{17},_{10} {Ne}^{17}

4

_{8} O^{15}, F^{17}

Explanation:

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.
$_{7} N^{15},_{8} O^{15}$
Same mass number $= 15$
Different atomic number 7, 8
$_{7} N^{15}$ and $_{8} O^{15}$ are mirror isobars because protons and neutrons are exchanged.

MP PET -2013

NUCLEAR PHYSICS

147496 The stable nucleus that has a radius half that of ${Fe}^{56}$ is

1

{Li}^{7}

2

{Na}^{21}

3

S^{16}

4

{Ca}^{40}

Explanation:

A Given, $R_{1} = R, R_{2} = R / 2, A_{1} = 56$
We know that,
$R = R_{0} A^{1 / 3}$
$R \propto A^{1 / 3}$
$A \propto R^{3}$
So, $\frac{A_{1}}{A_{2}} = {(\frac{R_{1}}{R_{2}})}^{1 / 3}$
$\frac{56}{A_{2}} = {(\frac{R}{R / 2})}^{3}$
$\frac{56}{A_{2}} = 8$
$A_{2} = \frac{56}{8} = 7$
The element with mass number 7 it means ${Li}^{7}$ .

AIPMT- 1997

NUCLEAR PHYSICS

147499 What is the particle $x$ in the following nuclear reaction?
$_{4}^{9} Be +_{2}^{4} He ⟶_{6}^{12} C + X$

1 Electron

2 Proton

3 Photon

4 Neutron

Explanation:

D Given the nuclear reaction $_{4} {Be}^{9} +_{2} {He}^{4} \to_{6} C^{12} +_{Z} X^{A}$
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 $_{0} n^{1}$ (neutron)
$Then, equation_{4} {Be}^{9} +_{2} {He}^{4} \to_{6} C^{12} +_{0} n^{1}$

WB JEE 2009

NUCLEAR PHYSICS

147500 An alpha nucleus of energy $\frac{1}{2} {mv}^{2}$ bombards a heavy nuclear target of charge $Ze$ . Then the distance of closest approach for the alpha nucleus will be proportional to

1

v^{2}

2

1 / m

3

1 / v^{4}

4

1 / Ze

Explanation:

B Initial kinetic energy of the particle (K.E. $)_{i} = \frac{1}{2} {mV}^{2}$
Also initial potential energy of the system $= 0$
Atomic number of alpha particle $Z_{α} = 2$
Final potential energy of the system at close
$(P.E.)_{f} = \frac{k (Ze) (2 e)}{r}$
According to the law of conservation of energy -
$(K.E.)_{i} + (P.E.)_{i} = (K.E.)_{f} + (P.E.)_{f}$
$\frac{1}{2} {mv}^{2} + 0 = \frac{0 + k 2 {Ze}^{2}}{r}$
$r = \frac{4 {kZe}^{2}}{{mv}^{2}}$
$r \propto \frac{1}{m} (Where, k = \frac{1}{4 π ε_{0}})$

JIPMER-2009

NUCLEAR PHYSICS

147507 Pick mirror isobars from the following:
$_{5} B^{12},_{6} C^{12},_{7} N^{15},_{8} O^{15},_{9} F^{17},_{10} {Ne}^{17}$

1

_{5} B^{12},_{6} C^{12}

2

_{7} N^{15},_{8} O^{15}

3

_{9} F^{17},_{10} {Ne}^{17}

4

_{8} O^{15}, F^{17}

Explanation:

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.
$_{7} N^{15},_{8} O^{15}$
Same mass number $= 15$
Different atomic number 7, 8
$_{7} N^{15}$ and $_{8} O^{15}$ are mirror isobars because protons and neutrons are exchanged.

MP PET -2013