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NUCLEAR PHYSICS

147639 When $_{90} {Th}^{228}$ transforms to $_{83} {Bi}^{212}$ , then the number to the emitted $α$ and $β$ -particles is, respectively

1

8 α, 7 β

2

4 α, 7 β

3

4 β, 7 β

4

4 α, 1 β

Explanation:

D $_{90} {Th}^{228} \to_{83} {Bi}^{212}$
Mass number is reduced by $= 228 - 212$
$= 16$
Number of alpha particle emitted are $(n_{α}) = \frac{16}{4} = 4$
Number of $β$ -particle emitted are $(n_{β})$ -
$= 2 n_{α} - Z + Z^{'}$
$= 2 \times 4 - 90 + 83$
$= 1$
4- $α$ particle and 1- $β$ particle are emitted.

Manipal UGET-2014

NUCLEAR PHYSICS

147640 The fraction of atoms of radioactive element that decays in 6 days is $7 / 8$ . The fraction that decays is 10 days will be

1

77 / 80

2

71 / 80

3

31 / 32

4

15 / 16

Explanation:

C Number of radioactive element that decays $= \frac{7}{8}$
Number of radioactive element that left decays $(\frac{N}{N_{o}}) =$
$= 1 - \frac{7}{8} = \frac{1}{8}$
According to radioactive decay law,
$N = N_{0} {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}}$
$\frac{N}{N_{0}} = {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}}$
$t = T_{1 / 2} \frac{\log e (\frac{N_{0}}{N})}{\log 2}$
$t = \frac{T_{1 / 2} \ln (N_{0} / N)}{\ln 2}$
$6 = \frac{T_{1 / 2} \ln (8 / 1)}{\ln 2}$
$10 = \frac{T_{1 / 2} \ln (N_{0} / N)}{\ln (2)}$
On dividing equation (i) and (ii), we get-
$\frac{6}{10} = \frac{\ln (8)}{\ln (N_{0} / N)}$
$\ln (\frac{N_{0}}{N}) = \frac{5}{3} \ln (8)$
$\ln (\frac{N_{0}}{N}) = \ln (8)^{5 / 3}$
$\ln (\frac{N_{0}}{N}) = \ln (32)$
$\frac{N_{0}}{N} = 32$
$\frac{N}{N_{0}} = \frac{1}{32}$
So, the fraction that decays is 10 days will be-
$= 1 - \frac{1}{32} = \frac{31}{32}$

Manipal UGET-2014

NUCLEAR PHYSICS

147641 A nucleus $X$ initially at rest, undergoes alpha decay according to the equation
$_{92} X^{A} \to_{Z} Y^{228} + α$
Then, the value of $A$ and $Z$ are

1 94,230

2 232,90

3 190,32

4 230,94

Explanation:

B The equation is given by,
$_{92} X^{A} \to_{Z} Y^{228} + α$
We can write above equation as-
$_{92} X^{A} \to_{Z} Y^{228} +_{2} {He}^{4} ∴ α =_{2} {He}^{4}$
Let atomic number and mass number is $Z$ and $A$ .
So,
$A = 228 + 4$
$A = 232$
And, $92 = Z + 2$
$Z = 90$
Hence, the value of $A$ and $Z$ are 232 and 90 .

Manipal UGET-2012

NUCLEAR PHYSICS

147642 A count rate meter shows a count of 240 per minute from a given radioactive source. One hour later the meter shows a count rate of 30 per minute. The half-life of the source is

1

80 \min

2

120 \min

3

20 \min

4

30 \min

Explanation:

C Given, $N_{o} = 240$ per minute $N = 30$ per minute
Time $(t) = 1 hr = 60$ minute
The equation for initial and final count rate is-
$N = N_{o} {(\frac{1}{2})}^{n}$
Where,
$N =$ final count rate
$N_{o} =$ initial count rate
$n =$ number of half lives
Then,
$30 = 240 {(\frac{1}{2})}^{\frac{60}{t_{1 / 2}}}$
$\frac{30}{240} = {(\frac{1}{2})}^{\frac{60}{t_{1 / 2}}}$
$\frac{1}{8} = {(\frac{1}{2})}^{\frac{60}{t_{1 / 2}}}$
${(\frac{1}{2})}^{3} = {(\frac{1}{2})}^{\frac{60}{t_{1 / 2}}}$
$\frac{60}{t_{1 / 2}} = 3$
$t_{1 / 2} = 20 \min$

Manipal UGET-2012

NUCLEAR PHYSICS

147639 When $_{90} {Th}^{228}$ transforms to $_{83} {Bi}^{212}$ , then the number to the emitted $α$ and $β$ -particles is, respectively

1

8 α, 7 β

2

4 α, 7 β

3

4 β, 7 β

4

4 α, 1 β

Explanation:

D $_{90} {Th}^{228} \to_{83} {Bi}^{212}$
Mass number is reduced by $= 228 - 212$
$= 16$
Number of alpha particle emitted are $(n_{α}) = \frac{16}{4} = 4$
Number of $β$ -particle emitted are $(n_{β})$ -
$= 2 n_{α} - Z + Z^{'}$
$= 2 \times 4 - 90 + 83$
$= 1$
4- $α$ particle and 1- $β$ particle are emitted.

Manipal UGET-2014

NUCLEAR PHYSICS

147640 The fraction of atoms of radioactive element that decays in 6 days is $7 / 8$ . The fraction that decays is 10 days will be

1

77 / 80

2

71 / 80

3

31 / 32

4

15 / 16

Explanation:

C Number of radioactive element that decays $= \frac{7}{8}$
Number of radioactive element that left decays $(\frac{N}{N_{o}}) =$
$= 1 - \frac{7}{8} = \frac{1}{8}$
According to radioactive decay law,
$N = N_{0} {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}}$
$\frac{N}{N_{0}} = {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}}$
$t = T_{1 / 2} \frac{\log e (\frac{N_{0}}{N})}{\log 2}$
$t = \frac{T_{1 / 2} \ln (N_{0} / N)}{\ln 2}$
$6 = \frac{T_{1 / 2} \ln (8 / 1)}{\ln 2}$
$10 = \frac{T_{1 / 2} \ln (N_{0} / N)}{\ln (2)}$
On dividing equation (i) and (ii), we get-
$\frac{6}{10} = \frac{\ln (8)}{\ln (N_{0} / N)}$
$\ln (\frac{N_{0}}{N}) = \frac{5}{3} \ln (8)$
$\ln (\frac{N_{0}}{N}) = \ln (8)^{5 / 3}$
$\ln (\frac{N_{0}}{N}) = \ln (32)$
$\frac{N_{0}}{N} = 32$
$\frac{N}{N_{0}} = \frac{1}{32}$
So, the fraction that decays is 10 days will be-
$= 1 - \frac{1}{32} = \frac{31}{32}$

Manipal UGET-2014

NUCLEAR PHYSICS

147641 A nucleus $X$ initially at rest, undergoes alpha decay according to the equation
$_{92} X^{A} \to_{Z} Y^{228} + α$
Then, the value of $A$ and $Z$ are

1 94,230

2 232,90

3 190,32

4 230,94

Explanation:

B The equation is given by,
$_{92} X^{A} \to_{Z} Y^{228} + α$
We can write above equation as-
$_{92} X^{A} \to_{Z} Y^{228} +_{2} {He}^{4} ∴ α =_{2} {He}^{4}$
Let atomic number and mass number is $Z$ and $A$ .
So,
$A = 228 + 4$
$A = 232$
And, $92 = Z + 2$
$Z = 90$
Hence, the value of $A$ and $Z$ are 232 and 90 .

Manipal UGET-2012

NUCLEAR PHYSICS

147642 A count rate meter shows a count of 240 per minute from a given radioactive source. One hour later the meter shows a count rate of 30 per minute. The half-life of the source is

1

80 \min

2

120 \min

3

20 \min

4

30 \min

Explanation:

C Given, $N_{o} = 240$ per minute $N = 30$ per minute
Time $(t) = 1 hr = 60$ minute
The equation for initial and final count rate is-
$N = N_{o} {(\frac{1}{2})}^{n}$
Where,
$N =$ final count rate
$N_{o} =$ initial count rate
$n =$ number of half lives
Then,
$30 = 240 {(\frac{1}{2})}^{\frac{60}{t_{1 / 2}}}$
$\frac{30}{240} = {(\frac{1}{2})}^{\frac{60}{t_{1 / 2}}}$
$\frac{1}{8} = {(\frac{1}{2})}^{\frac{60}{t_{1 / 2}}}$
${(\frac{1}{2})}^{3} = {(\frac{1}{2})}^{\frac{60}{t_{1 / 2}}}$
$\frac{60}{t_{1 / 2}} = 3$
$t_{1 / 2} = 20 \min$

Manipal UGET-2012

NUCLEAR PHYSICS

147639 When $_{90} {Th}^{228}$ transforms to $_{83} {Bi}^{212}$ , then the number to the emitted $α$ and $β$ -particles is, respectively

1

8 α, 7 β

2

4 α, 7 β

3

4 β, 7 β

4

4 α, 1 β

Explanation:

D $_{90} {Th}^{228} \to_{83} {Bi}^{212}$
Mass number is reduced by $= 228 - 212$
$= 16$
Number of alpha particle emitted are $(n_{α}) = \frac{16}{4} = 4$
Number of $β$ -particle emitted are $(n_{β})$ -
$= 2 n_{α} - Z + Z^{'}$
$= 2 \times 4 - 90 + 83$
$= 1$
4- $α$ particle and 1- $β$ particle are emitted.

Manipal UGET-2014

NUCLEAR PHYSICS

147640 The fraction of atoms of radioactive element that decays in 6 days is $7 / 8$ . The fraction that decays is 10 days will be

1

77 / 80

2

71 / 80

3

31 / 32

4

15 / 16

Explanation:

C Number of radioactive element that decays $= \frac{7}{8}$
Number of radioactive element that left decays $(\frac{N}{N_{o}}) =$
$= 1 - \frac{7}{8} = \frac{1}{8}$
According to radioactive decay law,
$N = N_{0} {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}}$
$\frac{N}{N_{0}} = {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}}$
$t = T_{1 / 2} \frac{\log e (\frac{N_{0}}{N})}{\log 2}$
$t = \frac{T_{1 / 2} \ln (N_{0} / N)}{\ln 2}$
$6 = \frac{T_{1 / 2} \ln (8 / 1)}{\ln 2}$
$10 = \frac{T_{1 / 2} \ln (N_{0} / N)}{\ln (2)}$
On dividing equation (i) and (ii), we get-
$\frac{6}{10} = \frac{\ln (8)}{\ln (N_{0} / N)}$
$\ln (\frac{N_{0}}{N}) = \frac{5}{3} \ln (8)$
$\ln (\frac{N_{0}}{N}) = \ln (8)^{5 / 3}$
$\ln (\frac{N_{0}}{N}) = \ln (32)$
$\frac{N_{0}}{N} = 32$
$\frac{N}{N_{0}} = \frac{1}{32}$
So, the fraction that decays is 10 days will be-
$= 1 - \frac{1}{32} = \frac{31}{32}$

Manipal UGET-2014

NUCLEAR PHYSICS

147641 A nucleus $X$ initially at rest, undergoes alpha decay according to the equation
$_{92} X^{A} \to_{Z} Y^{228} + α$
Then, the value of $A$ and $Z$ are

1 94,230

2 232,90

3 190,32

4 230,94

Explanation:

B The equation is given by,
$_{92} X^{A} \to_{Z} Y^{228} + α$
We can write above equation as-
$_{92} X^{A} \to_{Z} Y^{228} +_{2} {He}^{4} ∴ α =_{2} {He}^{4}$
Let atomic number and mass number is $Z$ and $A$ .
So,
$A = 228 + 4$
$A = 232$
And, $92 = Z + 2$
$Z = 90$
Hence, the value of $A$ and $Z$ are 232 and 90 .

Manipal UGET-2012

NUCLEAR PHYSICS

147642 A count rate meter shows a count of 240 per minute from a given radioactive source. One hour later the meter shows a count rate of 30 per minute. The half-life of the source is

1

80 \min

2

120 \min

3

20 \min

4

30 \min

Explanation:

C Given, $N_{o} = 240$ per minute $N = 30$ per minute
Time $(t) = 1 hr = 60$ minute
The equation for initial and final count rate is-
$N = N_{o} {(\frac{1}{2})}^{n}$
Where,
$N =$ final count rate
$N_{o} =$ initial count rate
$n =$ number of half lives
Then,
$30 = 240 {(\frac{1}{2})}^{\frac{60}{t_{1 / 2}}}$
$\frac{30}{240} = {(\frac{1}{2})}^{\frac{60}{t_{1 / 2}}}$
$\frac{1}{8} = {(\frac{1}{2})}^{\frac{60}{t_{1 / 2}}}$
${(\frac{1}{2})}^{3} = {(\frac{1}{2})}^{\frac{60}{t_{1 / 2}}}$
$\frac{60}{t_{1 / 2}} = 3$
$t_{1 / 2} = 20 \min$

Manipal UGET-2012

NUCLEAR PHYSICS

147639 When $_{90} {Th}^{228}$ transforms to $_{83} {Bi}^{212}$ , then the number to the emitted $α$ and $β$ -particles is, respectively

1

8 α, 7 β

2

4 α, 7 β

3

4 β, 7 β

4

4 α, 1 β

Explanation:

D $_{90} {Th}^{228} \to_{83} {Bi}^{212}$
Mass number is reduced by $= 228 - 212$
$= 16$
Number of alpha particle emitted are $(n_{α}) = \frac{16}{4} = 4$
Number of $β$ -particle emitted are $(n_{β})$ -
$= 2 n_{α} - Z + Z^{'}$
$= 2 \times 4 - 90 + 83$
$= 1$
4- $α$ particle and 1- $β$ particle are emitted.

Manipal UGET-2014

NUCLEAR PHYSICS

147640 The fraction of atoms of radioactive element that decays in 6 days is $7 / 8$ . The fraction that decays is 10 days will be

1

77 / 80

2

71 / 80

3

31 / 32

4

15 / 16

Explanation:

C Number of radioactive element that decays $= \frac{7}{8}$
Number of radioactive element that left decays $(\frac{N}{N_{o}}) =$
$= 1 - \frac{7}{8} = \frac{1}{8}$
According to radioactive decay law,
$N = N_{0} {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}}$
$\frac{N}{N_{0}} = {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}}$
$t = T_{1 / 2} \frac{\log e (\frac{N_{0}}{N})}{\log 2}$
$t = \frac{T_{1 / 2} \ln (N_{0} / N)}{\ln 2}$
$6 = \frac{T_{1 / 2} \ln (8 / 1)}{\ln 2}$
$10 = \frac{T_{1 / 2} \ln (N_{0} / N)}{\ln (2)}$
On dividing equation (i) and (ii), we get-
$\frac{6}{10} = \frac{\ln (8)}{\ln (N_{0} / N)}$
$\ln (\frac{N_{0}}{N}) = \frac{5}{3} \ln (8)$
$\ln (\frac{N_{0}}{N}) = \ln (8)^{5 / 3}$
$\ln (\frac{N_{0}}{N}) = \ln (32)$
$\frac{N_{0}}{N} = 32$
$\frac{N}{N_{0}} = \frac{1}{32}$
So, the fraction that decays is 10 days will be-
$= 1 - \frac{1}{32} = \frac{31}{32}$

Manipal UGET-2014

NUCLEAR PHYSICS

147641 A nucleus $X$ initially at rest, undergoes alpha decay according to the equation
$_{92} X^{A} \to_{Z} Y^{228} + α$
Then, the value of $A$ and $Z$ are

1 94,230

2 232,90

3 190,32

4 230,94

Explanation:

B The equation is given by,
$_{92} X^{A} \to_{Z} Y^{228} + α$
We can write above equation as-
$_{92} X^{A} \to_{Z} Y^{228} +_{2} {He}^{4} ∴ α =_{2} {He}^{4}$
Let atomic number and mass number is $Z$ and $A$ .
So,
$A = 228 + 4$
$A = 232$
And, $92 = Z + 2$
$Z = 90$
Hence, the value of $A$ and $Z$ are 232 and 90 .

Manipal UGET-2012

NUCLEAR PHYSICS

147642 A count rate meter shows a count of 240 per minute from a given radioactive source. One hour later the meter shows a count rate of 30 per minute. The half-life of the source is

1

80 \min

2

120 \min

3

20 \min

4

30 \min

Explanation:

C Given, $N_{o} = 240$ per minute $N = 30$ per minute
Time $(t) = 1 hr = 60$ minute
The equation for initial and final count rate is-
$N = N_{o} {(\frac{1}{2})}^{n}$
Where,
$N =$ final count rate
$N_{o} =$ initial count rate
$n =$ number of half lives
Then,
$30 = 240 {(\frac{1}{2})}^{\frac{60}{t_{1 / 2}}}$
$\frac{30}{240} = {(\frac{1}{2})}^{\frac{60}{t_{1 / 2}}}$
$\frac{1}{8} = {(\frac{1}{2})}^{\frac{60}{t_{1 / 2}}}$
${(\frac{1}{2})}^{3} = {(\frac{1}{2})}^{\frac{60}{t_{1 / 2}}}$
$\frac{60}{t_{1 / 2}} = 3$
$t_{1 / 2} = 20 \min$

Manipal UGET-2012