Physics Nuclei

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New answer posted•a year ago

Physics Nuclei Class 12th

Read the following statements:

(A) Volume of the nucleus is directly proportional to the mass number.

(B) Volume of the nucleus is independent of mass number.

(C) Density of the nucleus is directly proportional to the mass number.

(D) Density of nucleus is directly proportional to the cube root of the mass number.

(E) Density of the nucleus is independent of the mass number.

Choose the correct option from the following options.

3 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

Radius, R = R₀A^1/3

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New answer posted•a year ago

Physics Nuclei Nuclei

The activity of a radioactive sample falls from $700 s^{- 1}$ to $500 s^{- 1}$ in 30 minutes. Its half life is close to

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$A = A_{0} e^{- λ t}$

$\begin{array}{r} 500 = 700 e^{- λ t} \\ λ t = l n ? \frac{7}{5} \\ \frac{l n ? 2}{t_{1 / 2}} \times 30 = l n ? \frac{7}{5} \\ ? t_{1 / 2} = \frac{l n ? 2 \times 30}{l n ? \frac{7}{5}} \end{array}$

$t_{1 / 2} = 61.8 m i n \approx 62 m i n$

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New answer posted•a year ago

Physics Nuclei Class 12th

What is the half – life period of a radioactive material if its activity drops to 1/16^th of its initial value in 30 years?

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$A = \frac{A_{0}}{16} = \frac{A_{0}}{2^{30 / t_{1 / 2}}} \Rightarrow 2^{4} = 2^{30 / t_{1 / 2}}$

$\Rightarrow \frac{A_{0}}{16} = \frac{A_{0}}{2^{30 / t_{1 / 2}}} \Rightarrow 2^{4} = 2^{30 / t_{1 / 2}}$

$\Rightarrow t_{1 / 2} = \frac{30}{4} = 7.5$ years

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New answer posted•a year ago

Physics Nuclei Class 12th

Read the following statements:

(A) Volume of the nucleus is directly proportional to the mass number.

(B) Volume of the nucleus is independent of mass number.

(C) Density of the nucleus is directly proportional to the mass number.

(D) Density of nucleus is directly proportional to the cube root of the mass number.

(E) Density of the nucleus is independent of the mass number.

Choose the correct option from the following options

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

Radius, R = R₀A^1/3

Upvote

New answer posted•a year ago

Physics Nuclei Class 12th

The half life period of a radioactive substance is 60 days. The time taken for $\frac{7}{8} t h$  of its original mass to disintegrate will be:

2 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$t_{\frac{1}{2}} = 60 d a y s$

at t = 0 $\frac{m}{m_{0}} \to$ 0

$a t t \to \frac{m}{m_{0}} (1 - \frac{7}{8})$

= $\frac{m}{8 m_{0}} \to \frac{7}{8} (\frac{m}{m_{0}})$

$\Rightarrow \frac{1}{2^{3}} = \frac{1}{2^{(\frac{t}{60})}} \Rightarrow \frac{t}{60} = 3$

t = 180 days

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New answer posted•a year ago

Physics Nuclei Nuclei

Two radioactive materials A and B have decay constants 25 $λ$ and 16 $λ$ respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of B to that of A will be “e” after a time $\frac{1}{a λ} .$  The value of a is__________.

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$λ_{A} = 25 λ,$ $λ_{B} = 16 λ$

At t = 0 Þ N_A = N_B = N₀

after t = $\frac{1}{a λ} : - \frac{N_{B}}{N_{A}} = \frac{N_{0} e^{- 16 λ t}}{N_{0} e^{- 25 λ t}} = e^{(25 λ - 16 λ) t}$

->e = $e^{(9 λ \frac{1}{a λ})}$

$\Rightarrow \frac{9}{a} = 1 \Rightarrow a = 9$

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New answer posted•a year ago

Physics Nuclei Class 12th

Two radioactive materials A and B have decay constants 25 $λ$  and 16 $λ$ respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of B to that of A will be “e” after a time $\frac{1}{a λ} .$  The value of a is__________.

2 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

$λ_{A} = 25 λ,$ $λ_{B} = 16 λ$

At t = 0 N_A = N_B = N₀

after t = $\frac{1}{a λ} : - \frac{N_{B}}{N_{A}} = \frac{N_{0} e^{- 16 λ t}}{N_{0} e^{- 25 λ t}} = e^{(25 λ - 16 λ) t}$

e = $e^{(9 λ \frac{1}{a λ})}$

$\Rightarrow \frac{9}{a} = 1 \Rightarrow a = 9$

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New answer posted•a year ago

Physics Nuclei Class 12th

A sample contains 10^-2kg each of two substances A and B with half lives 4s and 8s respectively. The ration of their atomic weights is 1 : 2. The ratio of the amounts of A and B after16s is $\frac{x}{100}$ . The value of x is ______.

6 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

$f o r A, t_{\frac{1}{2}} = 4 s e c$

$= m A_{0} e^{-} \frac{0.693}{4} \times 16$

$m_{A} = m_{A 0} e^{- 4 \times 0.693}$ - (1)

for B,

for B, $t_{\frac{1}{2}} = 8 s e c$

$m_{B} = m_{B_{0}} e^{- 2 \times 0.693}$ - (2)

$\frac{m_{A}}{m_{B}} = \frac{m_{A O}}{m_{B O}} \frac{e^{- 4} \times 0.693}{e^{- 2} \times 0.693}$

$\frac{m_{A}}{m_{B}} = \frac{25}{100} = \frac{x}{100} x = 25$

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New answer posted•a year ago

Physics Nuclei Class 12th

Nucleus A is having mass number 220 and its binding energy per nucleon is 5.6 MeV. It splits in two fragments ‘B’ and ‘C’ of mass numbers 105 and 115. The binding energy of nucleons in (B) and ‘C’ 6.4 MeV per nucleon. The energy Q released per fission will be :

6 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$A \to^{105} B +^{115} C$

$\Rightarrow Q = [105 \times 6.4 + 115 \times 6.4] - [220 \times 5.6] M e v$

Q = 176 MeV

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New answer posted•a year ago

Physics Nuclei Class 12th

The half life of a radioactive substance is 5 years. After x years a given sample of the radioactive substance gets reduced to 6/25% of its initial value. The value of x is___________.

4 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

According to Nuclear activity, we can write

$N_{0} \overset{t_{\frac{1}{2}}}{\to} \frac{N_{0}}{2} \overset{t_{\frac{1}{2}}}{\to} \frac{N_{0}}{4} \overset{t_{\frac{1}{2}}}{\to} \frac{N_{0}}{8} \overset{t_{\frac{1}{2}}}{\to} \frac{N_{0}}{16} = (0.0625) N_{0}$

Time required = 4 × $t_{\frac{1}{2}} = 20 y r s$

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