The mass of a nucleus _Z^AX is less than the sum of the masses of (A-Z) number of neutrons and Z number of protons in the nucleus. The energy equivalent to the corresponding mass difference is known as the binding energy of the nucleus. A heavy nucleus of mass M can break into two light nuclei of masses m 1 and m 2 only if (m 1 + m 2 ) 3 and m 4 can undergo complete fusion and form a heavy nucleus of mass M' only if (m 3 + m 4 ) > M'. The masses of some neutral atoms are given in the table below : .tg border-collapse:collapse;border-spacing:0; .tg tdborder-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; overflow:hidden;padding:10px 5px;word-break:normal; .tg thborder-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal; .tg .tg-0laxtext-align:left;vertical-align:top _1^1H 1.007825 u _1^2H 2.014102 u _3^6Li 6.015123 u _3^7Li 7.016004 u _64^152Gd 151.919803 u _82^206Pb 205.974455 u _1^3H 3.016050 u _2^4He 4.002603 u _30^70Zn 69.925325 u _34^82Se 81.916709 u _83^209Bi 208.980388 u _84^210Po 209.982876 u (1 u = 932 MeV/c 2 ) The kinetic energy (in keV) of the alpha particle, when the nucleus _84^210Po at rest undergoes alpha decay, is

Atoms and Nuclei PYQ — JEE Advanced 2013 · Shift 2 · Q57 with Solution

The mass of a nucleus $_Z^AX$ is less than the sum of the masses of (A-Z) number of neutrons and Z number of protons in the nucleus. The energy equivalent to the corresponding mass difference is known as the binding energy of the nucleus. A heavy nucleus of mass M can break into two light nuclei of masses m₁ and m₂ only if (m₁ + m₂) < M. Also two light nuclei of masses m₃ and m₄ can undergo complete fusion and form a heavy nucleus of mass M' only if (m₃ + m₄) > M'. The masses of some neutral atoms are given in the table below :

$_1^1H$	1.007825 u	$_1^2H$	2.014102 u
$_3^6Li$	6.015123 u	$_3^7Li$	7.016004 u
$_{64}^{152}Gd$	151.919803 u	$_{82}^{206}Pb$	205.974455 u
$_1^3H$	3.016050 u	$_2^4He$	4.002603 u
$_{30}^{70}Zn$	69.925325 u	$_{34}^{82}Se$	81.916709 u
$_{83}^{209}Bi$	208.980388 u	$_{84}^{210}Po$	209.982876 u

(1 u = 932 MeV/c²)

The kinetic energy (in keV) of the alpha particle, when the nucleus

_{84}^{210}Po

at rest undergoes alpha decay, is

A
5319
B
5422
C
5707
D
5818

View written solutionFree

Correct answer: A

Write the decay equation

For alpha decay of polonium:

{}_{84}^{210}\text{Po} \to {}_{82}^{206}\text{Pb} + {}_{2}^{4}\text{He}

Since atomic masses are given, we can directly use them because the total number of electrons balances on both sides:

${}^{210}\text{Po}$ has 84 electrons
${}^{206}\text{Pb}$ has 82 electrons
${}^{4}\text{He}$ has 2 electrons

So electron masses cancel.

Calculate the Q-value of the decay

Given masses:

M({}^{210}\text{Po}) = 209.982876\,u

M({}^{206}\text{Pb}) = 205.974455\,u

M({}^{4}\text{He}) = 4.002603\,u

Mass defect:

\Delta m = M({}^{210}\text{Po}) - \left[M({}^{206}\text{Pb}) + M({}^{4}\text{He})\right]

\Delta m = 209.982876 - (205.974455 + 4.002603)

\Delta m = 209.982876 - 209.977058 = 0.005818\,u

Now,

Q = \Delta m\,c^2 = 0.005818 \times 932\,\text{MeV}

Q \approx 5.422\,\text{MeV}

Distribute the kinetic energy between daughter nucleus and alpha particle

Initial nucleus is at rest, so momentum conservation gives equal and opposite momenta for Pb and $\alpha$ .

For two-body decay:

K_\alpha = Q\cdot \frac{M_{\text{daughter}}}{M_{\text{daughter}} + M_\alpha}

Here,

K_\alpha = Q\cdot \frac{206}{206+4} = Q\cdot \frac{206}{210}

So,

K_\alpha = 5.422 \times \frac{206}{210}

K_\alpha \approx 5.422 \times 0.98095 \approx 5.319\,\text{MeV}

Convert to keV

5.319\,\text{MeV} = 5319\,\text{keV}

Check options

The correct option is:

\boxed{5319\,\text{keV}}

which corresponds to Option A.

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$_1^1H$	1.007825 u	$_1^2H$	2.014102 u
$_3^6Li$	6.015123 u	$_3^7Li$	7.016004 u
$_{64}^{152}Gd$	151.919803 u	$_{82}^{206}Pb$	205.974455 u
$_1^3H$	3.016050 u	$_2^4He$	4.002603 u
$_{30}^{70}Zn$	69.925325 u	$_{34}^{82}Se$	81.916709 u
$_{83}^{209}Bi$	208.980388 u	$_{84}^{210}Po$	209.982876 u

(1 u = 932 MeV/c²)

The kinetic energy (in keV) of the alpha particle, when the nucleus

_{84}^{210}Po

at rest undergoes alpha decay, is

A
5319
B
5422
C
5707
D
5818

View written solutionFree

Correct answer: A

Write the decay equation

For alpha decay of polonium:

{}_{84}^{210}\text{Po} \to {}_{82}^{206}\text{Pb} + {}_{2}^{4}\text{He}

Since atomic masses are given, we can directly use them because the total number of electrons balances on both sides:

${}^{210}\text{Po}$ has 84 electrons
${}^{206}\text{Pb}$ has 82 electrons
${}^{4}\text{He}$ has 2 electrons

So electron masses cancel.

Calculate the Q-value of the decay

Given masses:

M({}^{210}\text{Po}) = 209.982876\,u

M({}^{206}\text{Pb}) = 205.974455\,u

M({}^{4}\text{He}) = 4.002603\,u

Mass defect:

\Delta m = M({}^{210}\text{Po}) - \left[M({}^{206}\text{Pb}) + M({}^{4}\text{He})\right]

\Delta m = 209.982876 - (205.974455 + 4.002603)

\Delta m = 209.982876 - 209.977058 = 0.005818\,u

Now,

Q = \Delta m\,c^2 = 0.005818 \times 932\,\text{MeV}

Q \approx 5.422\,\text{MeV}

Distribute the kinetic energy between daughter nucleus and alpha particle

Initial nucleus is at rest, so momentum conservation gives equal and opposite momenta for Pb and $\alpha$ .

For two-body decay:

K_\alpha = Q\cdot \frac{M_{\text{daughter}}}{M_{\text{daughter}} + M_\alpha}

Here,

K_\alpha = Q\cdot \frac{206}{206+4} = Q\cdot \frac{206}{210}

So,

K_\alpha = 5.422 \times \frac{206}{210}

K_\alpha \approx 5.422 \times 0.98095 \approx 5.319\,\text{MeV}

Convert to keV

5.319\,\text{MeV} = 5319\,\text{keV}

Check options

The correct option is:

\boxed{5319\,\text{keV}}

which corresponds to Option A.

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