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Physics NCERT Exemplar Solutions Class 12th Chapter Thirteen

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3 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Thirteen Nuclei

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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3 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Thirteen Nuclei

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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3 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Thirteen Kinetic Theory

At a certain temperature, the degrees of freedom per molecule for gas is 8. The gas performs 150 J at work when it expands under constant pressure. The amount of heat absorbed by the gas will be________J.

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Vishal Baghel

Contributor-Level 10

Degree of freedom, f = 8

$∴ C_{V} = \frac{8}{2} R = 4 R a n d C_{P} = 4 R + R = 5 R$

So, H = $Δ U + W$

at constant pressure, W = $P Δ v$

150 = $n R Δ T$

So, $H = n C_{V} Δ T = n 5 R Δ T = 5 (n R Δ T)$

= 5 * 150

= 750 J

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3 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Thirteen Motion in a Straight Line

A ball is thrown vertically upwards with a velocity of 19.6 ms^-1 from the top of a tower. The ball strikes the ground after 6s. The height from the ground up to which the ball can rise will be $(\frac{k}{5}) m .$ The value of k is_______(use g = 9.8m/s²)

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Vishal Baghel

Contributor-Level 10

h = 39.2 + 19.6 = 58.8 m

Height above tower

$h' = \frac{u^{2}}{2 g} = \frac{19.6 * 19.6}{2 * 9.8}$

= 19.6

$A s, \frac{k}{5} = 78.4 \Rightarrow k = 392$

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3 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Thirteen Nuclei

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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3 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Thirteen Nuclei

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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3 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Thirteen Nuclei

The X-Y plane be taken as the boundary between two transparent media M₁ and M₂. M₁ in $z \geq 0$ has a refractive index of $\sqrt[]{2}$ and M₂ with Z < 0 has refractive index of $\sqrt[]{3} .$ A ray of light travelling in M₁ along the direction given by the vector $\vec{P} = 4 \sqrt[]{3} \hat{i} - 3 \sqrt[]{3} \hat{j} - 5 \hat{k},$ is incident on the plane of separation. The value of difference between the angle of incident in M₁ and the angle of refraction in M₂ will be_______ degree.

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Vishal Baghel

Contributor-Level 10

For angle of incidence 'i' :-

cos I = $\frac{5}{\sqrt[]{48 + 27 + 25}} = \frac{5}{\sqrt[]{100}}$

i = 60°

Using snell's law :-

$μ_{1} s i n i = μ_{2} s i n r$

sin r = $\frac{\sqrt[]{2}}{\sqrt[]{3}} s i n 60 ° = \sqrt[]{\frac{2}{3}} * \frac{\sqrt[]{3}}{2} = \frac{1}{\sqrt[]{2}}$

$∴ r = 45 °$

So, difference, I – r = 60° - 452 = 15°

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3 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Thirteen Nuclei

Find the binding energy per neucleon for $_{50}^{120} S n$ . Mass of proton $m_{p} = (A) 1.00783 U$ , mass of neutron $m_{n} = (A) 1.00867 U$ and mass of tin nucleus $m_{S n} = 119.902199 U$ . (take $1 U = 931 M e V$ )

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alok kumar singh

Contributor-Level 10

Sol. $\Rightarrow {\overset{?}{V}}_{1} = 2 \overset{?}{U_{c m}} - \overset{?}{U_{1}} \Rightarrow 2 [\frac{m / 2 V_{o}}{\frac{m}{2} + \frac{m}{3}}] - V_{o}$

$\Rightarrow \frac{6}{5} V_{o} - V_{o} \Rightarrow \frac{V_{0}}{5}$

$\Rightarrow λ_{o} = \frac{h c}{(\frac{m}{2} V_{o})} λ_{f} = \frac{h c}{(\frac{M}{2} \frac{V_{o}}{5})}$

$\Rightarrow Δ λ = \frac{8 h c}{{m V}_{o}}$

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3 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Thirteen Kinetic Theory

According to kinetic theory of gases,

A. The motion of the gas molecules freezes at 0°C.

B. The mean free path of gas molecules decreases if the density of molecules is increased.

C. The mean free path of gas molecules increases if temperature is increased keeping pressure constant.

D. Average kinetic energy per molecules per degree of freedom is $\frac{3}{2} k_{B} T$ (for monoatomic gases).

Choose the most appropriate answer form the options given below

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Payal Gupta

Contributor-Level 10

$\bar{x}$ (mean free path) = $\frac{1}{\sqrt[]{2} π d^{2} n_{v}}$

$n_{v} = \frac{n N_{A}}{v},$ n → No. of moles in volume v N_A ® Avogadro's Number

$\frac{n}{v} = \frac{P}{R_{T}}$

$\bar{x} \propto v \propto \frac{1}{ρ (d e n s i t y} & \bar{x} α T a t c o n s t a n t P)$

^{$ρ ↑ \Leftrightarrow \bar{x} ↓ | \bar{x} ↑ \Leftrightarrow T ↑$}

The motion of the gas molecules freezes at 0K not 0° C

Average kinetic Energy per molecule per degree of freedom is $= \frac{1}{2} k_{B} T$ (for Mono atomic gases)

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4 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Thirteen NEET

What is the difference between alpha, beta, and gamma decay?

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Pallavi Pathak

Contributor-Level 10

These are the difference between the alpha, beta and gamma decay:

Alpha decay: It reduces mass number by 4 and atomic number by 2. It is the emission of a helium nucleus (2 protons, 2 neutrons).
Beta decay: It changes atomic number by ±1 but there is no change in the mass number. It involves a neutron converting into a proton (or vice versa).
Gamma decay: Here, there is no change in the atomic or mass number, just the emission of high-energy photons.

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