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2 months ago

Nuclei physics ncert exemplar solutions class 12th chapter two

A radioactive nucleus can decay by two different processes. Half-life for the first process is 3.0 hours while it is 4.5 hours for the second process. The effective half-life of the nucleus will be:

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Vishal Baghel

Contributor-Level 10

$\frac{1}{t} = \frac{1}{t_{1}} + \frac{1}{t_{2}}$

$\frac{1}{t} = \frac{1}{3} = \frac{1}{4.5}$

$= \frac{4.5 + 3}{3 * 4.5}$

t = $\frac{3 * 4.5}{7.5} = \frac{9}{5}$

t = 1.8 hrs

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2 months ago

Nuclei Physics NCERT Exemplar Solutions Class 12th Chapter Thirteen

The magnetic flux through a coil perpendicular to its plane is varying according to the relation $ϕ = (5 t^{3} + 4 t^{2} + 2 t - 5)$ Weber. If the resistance of the coil is 5 ohm, then the induced current through the coil at t = 2s will be

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Payal Gupta

Contributor-Level 10

$V = \frac{- d ?}{d t} = - (15 t^{2} + 8 t + 2)$

at t = 2 sec, V = (15 * 4 + 8 * 2 + 2)

= 78 V

$I = \frac{V}{R} = \frac{78}{5} = 15.6 A$

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2 months ago

Nuclei Physics NCERT Exemplar Solutions Class 12th Chapter Thirteen

The activity of a radioactive material is 2.56 * 10^-3 Ci. If the half life of the material is 5 days, after how many days the activity will become 2 * 10^-5 Ci?

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Payal Gupta

Contributor-Level 10

A₀ = 2.56 * 10^-3C_i

A = 2 * 10^-15C_i

A = $\frac{A_{0}}{2^{n}}$

$\Rightarrow 2 * 1 0^{- 5} = \frac{2.56 * 1 0^{- 3}}{2^{n}}$

2ⁿ = 128

n = 7

t = 7 * t_1/2

= 7 * 5 = 35 days

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2 months ago

Nuclei Physics NCERT Exemplar Solutions Class 12th Chapter Thirteen

In the hydrogen spectrum, $λ$ be the wavelength of first transition line of Lyman series. The wavelength difference will be $" a λ "$ between the wavelength of 3rd transition line of Paschen series and that 2nd transition line B Balmer series where a = __________.

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Vishal Baghel

Contributor-Level 10

For first line of Lyman

$\frac{1}{λ} = R (1 - \frac{1}{4}) = R \frac{3}{4}$

$λ = \frac{4}{3 R}$

3rd line (Paschen)

$\frac{1}{λ_{3}} = R (\frac{1}{3^{2}} - \frac{1}{6^{2}}) = \frac{R}{9} * \frac{3}{4}$

2nd line (B almer)

${\frac{1}{λ}}_{2} = R (\frac{1}{2^{2}} - \frac{1}{4^{2}}) = \frac{R}{4} * \frac{3}{4}$

$a (\frac{4}{3 R}) = \frac{20}{3 R} \Rightarrow a = 5$

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2 months ago

Nuclei Physics NCERT Exemplar Solutions Class 12th Chapter Thirteen

Mass numbers of two nuclei are in the ratio of 4 : 3. Their nuclear densities will be in the ratio of

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alok kumar singh

Contributor-Level 10

$R = R ? A^{1 / 3}$

$δ = \frac{m a s s}{v o l u m e} = \frac{m A}{\frac{4}{3} π R^{3}} = \frac{m A}{4 / 3 π R_{0}^{3} A}$

$\Rightarrow$ Density is independent of mass number

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2 months ago

Nuclei Physics NCERT Exemplar Solutions Class 12th Chapter Thirteen

A nucleus of mass M at rest splits into two parts having masses $\frac{M'}{3}$ and $\frac{2 M'}{2} (M' < M),$ The ratio of de Broglie wavelength of two parts will be:

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alok kumar singh

Contributor-Level 10

$\frac{λ_{1}}{λ_{2}} = \frac{h / P_{1}}{h / P_{2}} = \frac{1}{1}$

P1 = P2 as Fnet = 0

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2 months ago

Nuclei Physics NCERT Exemplar Solutions Class 12th Chapter Thirteen

The disintegration rate of a certain radioactive sample at any instant is 4250 disintegrations per minutes later, the rate becomes 2250 disintegrations per minute. The approximate decay constant is:

(Take log₁₀ 1.88 = 0.274)

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Vishal Baghel

Contributor-Level 10

$A = A o e^{- λ t}$

$\Rightarrow 2250 = 4250 e^{- λ t}$

$\Rightarrow λ * 10 * 0.434 = 0.274$

$\Rightarrow λ = \frac{0.274}{10 * 0.434} = 0.63 m i n^{- 1}$

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2 months ago

Nuclei Physics NCERT Exemplar Solutions Class 12th Chapter Thirteen

A radioactive simple is undergoing decay. At any time t₁, its activity is A and another time t₂, the activity is $\frac{A}{5} .$ What is the average life time for the sample?

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Payal Gupta

Contributor-Level 10

$A = A_{0} e^{- λ t_{1}}$ [Radio active decay law]

$\frac{A}{5} = A_{0} e^{- λ (t_{2} - t_{1})}$

$\Rightarrow l n 5 = λ (t_{2} - t_{1})$

$\Rightarrow A v e r a g e l i f e = \frac{1}{λ} = \frac{(t_{2} - t_{1})}{l n 5}$

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2 months ago

Nuclei Physics NCERT Exemplar Solutions Class 12th Chapter Thirteen

A radioactive sample decays $\frac{7}{8}$ times its original quantity in 15 minutes. The half-life of the sample is

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Vishal Baghel

Contributor-Level 10

$N_{0} \to_{}^{t = 15 m i n} (N_{0} - \frac{7}{8} N_{0}) = \frac{N_{0}}{8}$

So, $N = \frac{N_{0}}{2^{t / t_{1 / 2}}} \Rightarrow \frac{N_{0}}{8} = \frac{N_{0}}{2^{t / t_{1 / 2}}}$

$\Rightarrow 2^{3} = 2^{(\frac{15 m i n}{T})}$

$\Rightarrow T = \frac{15 m i n}{3} = 5 m i n$

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2 months ago

Nuclei Physics NCERT Exemplar Solutions Class 12th Chapter Thirteen

The Q-value of a nuclear reaction and kinetic energy of the projectile particle, K_p are related as

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Payal Gupta

Contributor-Level 10

X + p → Y + b

$Q = K_{Y} + K_{b} - K_{P}$

$Q + K_{P} = K_{Y} + K_{b}$

$\Rightarrow Q + K_{P} > 0$

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