QBManager

PHXII13:NUCLEI

363901 $_{92} U^{235}$ undergoes successive disintegrations with end product of $_{82} P^{203}$ . The number of $α$ and $β$ particles emitted are

1

α = 8, β = 6

2

α = 3, β = 3

3

α = 6, β = 4

4

α = 6, β = 0

Explanation:

$_{92} U^{235}$ disintegration end product is $_{82} P^{203}$ with $α$ and $β$ particles emitted.
$Δ A = 235 - 203 = 32$
$∴$ 8 alpha particles are emitted. The charge ( $Z$ ) should be $92 - 16 = 76$
But as the final charge is 82, six $β^{-}$ particles are also emitted to make the final atomic number $Z = 82$ .
$∴$ 8 alpha particles and six $β^{-}$ are emitted

KCET - 2008

PHXII13:NUCLEI

363902 We have seen that a gamma ray dose of $3 G y$ is lethal to half the people exposed to it. If the equivalent energy were absorbed as heat, what rise in body temperature would result?

1

300 μ K

2

700 μ K

3

455 μ K

4

390 μ K

Explanation:

We can relate an absorbed energy $Q$ and the resulting temperature increase $Δ T$ with relation $Q = m r (Δ T)$ . In this equation, $m$ is the mass of the material absorbing the energy and $r$ is the specific heat of that material. An absorbed dose of $3 G y$ corresponds to an absorbed energy per unit mass of $3 J k g^{- 1}$ . Let us assume that $r$ is the specific heat of human body, is the same as that of water,
$4180 J k g^{- 1} K^{- 1}$ .
Then, we find that
$Δ T = \frac{Q / m}{r} = \frac{3}{4180}$
$= 7.2 \times 10^{- 4} K ≃ 700 μ K$

AIIMS - 2007

PHXII13:NUCLEI

363903 A radioactive nucleus emits $4 α$ - particles and $7 β$ - particles in succession. The ratio of number of neutrons of that of protons, is [ $A =$ mass number, $Z =$ atomic number]

1

\frac{A - Z - 13}{Z - 2}

2

\frac{A - Z - 15}{Z - 1}

3

\frac{A - Z - 13}{Z - 1}

4

\frac{A - Z - 11}{Z - 2}

Explanation:

Let us assume, a particle $X$ having atomic number $Z$ and mass number $A$
when an $α$ - particle is emitted by a nucleus, then its atomic number decreases by 2 and mass number decreases by 4. So, for given case
$_{Z} X^{A} {\overset{4 α - p a r t i c l e}{\to}}_{Z - 8} Y^{A - 16}$
When an $β$ -particle is emitted by a nucleus its atomic number increases by one and mass number remains unchanged. So, for given case
$_{Z - 8} Y^{A - 16} {\overset{7 β - p a r t i c l e}{\to}}_{Z - 1} Z^{A - 16}$
$∴ \frac{N u m b e r o f n e u t r o n s}{N u m b e r o f p r o t o n s} = \frac{(A - 16) - (Z - 1)}{(Z - 1)}$
$= \frac{A - Z - 15}{Z - 1}$

MHTCET - 2020

PHXII13:NUCLEI

363904 In which of the following processes, the number of protons in the nucleus increase?

1

α

-decay

2

β^{-}

-decay

3

β^{+}

-decay

4

k -

capture

Explanation:

For $α - d e c a y :_{x} A^{y} \to_{x - 2} B^{y - 4} + α$
For $β^{-} :_{x} A^{y} \to_{x + 1} B^{y} +_{- 1} β^{0}$
For $β^{+} :_{x} A^{y} \to_{x - 1} B^{y} +_{+ 1} β^{0}$
For $k$ -cpature, there will be no change in the number of protons. Hence, in $β$ decay.

PHXII13:NUCLEI

363901 $_{92} U^{235}$ undergoes successive disintegrations with end product of $_{82} P^{203}$ . The number of $α$ and $β$ particles emitted are

1

α = 8, β = 6

2

α = 3, β = 3

3

α = 6, β = 4

4

α = 6, β = 0

Explanation:

$_{92} U^{235}$ disintegration end product is $_{82} P^{203}$ with $α$ and $β$ particles emitted.
$Δ A = 235 - 203 = 32$
$∴$ 8 alpha particles are emitted. The charge ( $Z$ ) should be $92 - 16 = 76$
But as the final charge is 82, six $β^{-}$ particles are also emitted to make the final atomic number $Z = 82$ .
$∴$ 8 alpha particles and six $β^{-}$ are emitted

KCET - 2008

PHXII13:NUCLEI

363902 We have seen that a gamma ray dose of $3 G y$ is lethal to half the people exposed to it. If the equivalent energy were absorbed as heat, what rise in body temperature would result?

1

300 μ K

2

700 μ K

3

455 μ K

4

390 μ K

Explanation:

We can relate an absorbed energy $Q$ and the resulting temperature increase $Δ T$ with relation $Q = m r (Δ T)$ . In this equation, $m$ is the mass of the material absorbing the energy and $r$ is the specific heat of that material. An absorbed dose of $3 G y$ corresponds to an absorbed energy per unit mass of $3 J k g^{- 1}$ . Let us assume that $r$ is the specific heat of human body, is the same as that of water,
$4180 J k g^{- 1} K^{- 1}$ .
Then, we find that
$Δ T = \frac{Q / m}{r} = \frac{3}{4180}$
$= 7.2 \times 10^{- 4} K ≃ 700 μ K$

AIIMS - 2007

PHXII13:NUCLEI

363903 A radioactive nucleus emits $4 α$ - particles and $7 β$ - particles in succession. The ratio of number of neutrons of that of protons, is [ $A =$ mass number, $Z =$ atomic number]

1

\frac{A - Z - 13}{Z - 2}

2

\frac{A - Z - 15}{Z - 1}

3

\frac{A - Z - 13}{Z - 1}

4

\frac{A - Z - 11}{Z - 2}

Explanation:

Let us assume, a particle $X$ having atomic number $Z$ and mass number $A$
when an $α$ - particle is emitted by a nucleus, then its atomic number decreases by 2 and mass number decreases by 4. So, for given case
$_{Z} X^{A} {\overset{4 α - p a r t i c l e}{\to}}_{Z - 8} Y^{A - 16}$
When an $β$ -particle is emitted by a nucleus its atomic number increases by one and mass number remains unchanged. So, for given case
$_{Z - 8} Y^{A - 16} {\overset{7 β - p a r t i c l e}{\to}}_{Z - 1} Z^{A - 16}$
$∴ \frac{N u m b e r o f n e u t r o n s}{N u m b e r o f p r o t o n s} = \frac{(A - 16) - (Z - 1)}{(Z - 1)}$
$= \frac{A - Z - 15}{Z - 1}$

MHTCET - 2020

PHXII13:NUCLEI

363904 In which of the following processes, the number of protons in the nucleus increase?

1

α

-decay

2

β^{-}

-decay

3

β^{+}

-decay

4

k -

capture

Explanation:

For $α - d e c a y :_{x} A^{y} \to_{x - 2} B^{y - 4} + α$
For $β^{-} :_{x} A^{y} \to_{x + 1} B^{y} +_{- 1} β^{0}$
For $β^{+} :_{x} A^{y} \to_{x - 1} B^{y} +_{+ 1} β^{0}$
For $k$ -cpature, there will be no change in the number of protons. Hence, in $β$ decay.

PHXII13:NUCLEI

363901 $_{92} U^{235}$ undergoes successive disintegrations with end product of $_{82} P^{203}$ . The number of $α$ and $β$ particles emitted are

1

α = 8, β = 6

2

α = 3, β = 3

3

α = 6, β = 4

4

α = 6, β = 0

Explanation:

$_{92} U^{235}$ disintegration end product is $_{82} P^{203}$ with $α$ and $β$ particles emitted.
$Δ A = 235 - 203 = 32$
$∴$ 8 alpha particles are emitted. The charge ( $Z$ ) should be $92 - 16 = 76$
But as the final charge is 82, six $β^{-}$ particles are also emitted to make the final atomic number $Z = 82$ .
$∴$ 8 alpha particles and six $β^{-}$ are emitted

KCET - 2008

PHXII13:NUCLEI

363902 We have seen that a gamma ray dose of $3 G y$ is lethal to half the people exposed to it. If the equivalent energy were absorbed as heat, what rise in body temperature would result?

1

300 μ K

2

700 μ K

3

455 μ K

4

390 μ K

Explanation:

We can relate an absorbed energy $Q$ and the resulting temperature increase $Δ T$ with relation $Q = m r (Δ T)$ . In this equation, $m$ is the mass of the material absorbing the energy and $r$ is the specific heat of that material. An absorbed dose of $3 G y$ corresponds to an absorbed energy per unit mass of $3 J k g^{- 1}$ . Let us assume that $r$ is the specific heat of human body, is the same as that of water,
$4180 J k g^{- 1} K^{- 1}$ .
Then, we find that
$Δ T = \frac{Q / m}{r} = \frac{3}{4180}$
$= 7.2 \times 10^{- 4} K ≃ 700 μ K$

AIIMS - 2007

PHXII13:NUCLEI

363903 A radioactive nucleus emits $4 α$ - particles and $7 β$ - particles in succession. The ratio of number of neutrons of that of protons, is [ $A =$ mass number, $Z =$ atomic number]

1

\frac{A - Z - 13}{Z - 2}

2

\frac{A - Z - 15}{Z - 1}

3

\frac{A - Z - 13}{Z - 1}

4

\frac{A - Z - 11}{Z - 2}

Explanation:

Let us assume, a particle $X$ having atomic number $Z$ and mass number $A$
when an $α$ - particle is emitted by a nucleus, then its atomic number decreases by 2 and mass number decreases by 4. So, for given case
$_{Z} X^{A} {\overset{4 α - p a r t i c l e}{\to}}_{Z - 8} Y^{A - 16}$
When an $β$ -particle is emitted by a nucleus its atomic number increases by one and mass number remains unchanged. So, for given case
$_{Z - 8} Y^{A - 16} {\overset{7 β - p a r t i c l e}{\to}}_{Z - 1} Z^{A - 16}$
$∴ \frac{N u m b e r o f n e u t r o n s}{N u m b e r o f p r o t o n s} = \frac{(A - 16) - (Z - 1)}{(Z - 1)}$
$= \frac{A - Z - 15}{Z - 1}$

MHTCET - 2020

PHXII13:NUCLEI

363904 In which of the following processes, the number of protons in the nucleus increase?

1

α

-decay

2

β^{-}

-decay

3

β^{+}

-decay

4

k -

capture

Explanation:

For $α - d e c a y :_{x} A^{y} \to_{x - 2} B^{y - 4} + α$
For $β^{-} :_{x} A^{y} \to_{x + 1} B^{y} +_{- 1} β^{0}$
For $β^{+} :_{x} A^{y} \to_{x - 1} B^{y} +_{+ 1} β^{0}$
For $k$ -cpature, there will be no change in the number of protons. Hence, in $β$ decay.

PHXII13:NUCLEI

363901 $_{92} U^{235}$ undergoes successive disintegrations with end product of $_{82} P^{203}$ . The number of $α$ and $β$ particles emitted are

1

α = 8, β = 6

2

α = 3, β = 3

3

α = 6, β = 4

4

α = 6, β = 0

Explanation:

$_{92} U^{235}$ disintegration end product is $_{82} P^{203}$ with $α$ and $β$ particles emitted.
$Δ A = 235 - 203 = 32$
$∴$ 8 alpha particles are emitted. The charge ( $Z$ ) should be $92 - 16 = 76$
But as the final charge is 82, six $β^{-}$ particles are also emitted to make the final atomic number $Z = 82$ .
$∴$ 8 alpha particles and six $β^{-}$ are emitted

KCET - 2008

PHXII13:NUCLEI

363902 We have seen that a gamma ray dose of $3 G y$ is lethal to half the people exposed to it. If the equivalent energy were absorbed as heat, what rise in body temperature would result?

1

300 μ K

2

700 μ K

3

455 μ K

4

390 μ K

Explanation:

We can relate an absorbed energy $Q$ and the resulting temperature increase $Δ T$ with relation $Q = m r (Δ T)$ . In this equation, $m$ is the mass of the material absorbing the energy and $r$ is the specific heat of that material. An absorbed dose of $3 G y$ corresponds to an absorbed energy per unit mass of $3 J k g^{- 1}$ . Let us assume that $r$ is the specific heat of human body, is the same as that of water,
$4180 J k g^{- 1} K^{- 1}$ .
Then, we find that
$Δ T = \frac{Q / m}{r} = \frac{3}{4180}$
$= 7.2 \times 10^{- 4} K ≃ 700 μ K$

AIIMS - 2007

PHXII13:NUCLEI

363903 A radioactive nucleus emits $4 α$ - particles and $7 β$ - particles in succession. The ratio of number of neutrons of that of protons, is [ $A =$ mass number, $Z =$ atomic number]

1

\frac{A - Z - 13}{Z - 2}

2

\frac{A - Z - 15}{Z - 1}

3

\frac{A - Z - 13}{Z - 1}

4

\frac{A - Z - 11}{Z - 2}

Explanation:

Let us assume, a particle $X$ having atomic number $Z$ and mass number $A$
when an $α$ - particle is emitted by a nucleus, then its atomic number decreases by 2 and mass number decreases by 4. So, for given case
$_{Z} X^{A} {\overset{4 α - p a r t i c l e}{\to}}_{Z - 8} Y^{A - 16}$
When an $β$ -particle is emitted by a nucleus its atomic number increases by one and mass number remains unchanged. So, for given case
$_{Z - 8} Y^{A - 16} {\overset{7 β - p a r t i c l e}{\to}}_{Z - 1} Z^{A - 16}$
$∴ \frac{N u m b e r o f n e u t r o n s}{N u m b e r o f p r o t o n s} = \frac{(A - 16) - (Z - 1)}{(Z - 1)}$
$= \frac{A - Z - 15}{Z - 1}$

MHTCET - 2020

PHXII13:NUCLEI

363904 In which of the following processes, the number of protons in the nucleus increase?

1

α

-decay

2

β^{-}

-decay

3

β^{+}

-decay

4

k -

capture

Explanation:

For $α - d e c a y :_{x} A^{y} \to_{x - 2} B^{y - 4} + α$
For $β^{-} :_{x} A^{y} \to_{x + 1} B^{y} +_{- 1} β^{0}$
For $β^{+} :_{x} A^{y} \to_{x - 1} B^{y} +_{+ 1} β^{0}$
For $k$ -cpature, there will be no change in the number of protons. Hence, in $β$ decay.

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