Energy, \(E=10^{-3} \times\left(3 \times 10^{8}\right)^{2} \mathrm{~J}\)
\(E=10^{-3} \times 9 \times 10^{16}=9 \times 10^{13} \mathrm{~J}\)
Thus, if one gram of matter is converted to energy, there is a release of enormous amount of energy.

Yes, the instantaneous power can be negative as, \(P_{\text {instantaneous }}=I_{\text {in }} \times V_{\text {in }}=I_0 \sin \omega t \times V_0 \cos \omega t\) No, because it is average, so it will be positive.

The metal detector works on the principle of resonance in ac circuits. When you walk through a metal detector, you are, in fact, walking through a coil of many turns. The coil is connected to a capacitor tuned so that the circuit is in resonance. When you walk through with metal in your pocket, the impedance of the circuit changes – resulting in significant change in current in the circuit. This change in current is detected and the electronic circuitry causes a sound to be emitted as an alarm.

Here v = 0.5 Hz; N = 100, A = 0.1 m² and B = 0.01 T. Employing Equation.
\(\varepsilon_0=N B A(2 \pi v)\)
= 100 \(\times\)0.01 \(\times\)0.1 \(\times\) 2 \(\times\) 3.14 \(\times\) 0.5
= 0.314 V
The maximum voltage is 0.314 V.
We urge you to explore such alternative possibilities for power generation.

10⁶ms^-1

In 5700 years, the number of carbon atoms reduces to 50% of its initial count.
If R_X and R_Y are the rates of disintegration of nuclei X and Y, then
\( \frac{R_{X}}{R_{Y}}=\frac{N_{X}}{N_{Y}} \times \frac{\lambda_{X}}{\lambda_{Y}} \quad[\because R=\lambda N] \)
\(\frac{\lambda_{X}}{\lambda_{Y}}=\frac{T_{Y}}{T_{X}}=\frac{2}{1} \text { (given) }\left[\because T=\frac{0.693}{\lambda}\right] \)
After two hours,
\(\frac{N_{X}}{N_{Y}}=\frac{1}{2} \Rightarrow \frac{R_{X}}{R_{Y}}=1: 1\)

For a dynamically stable orbit in a hydrogen atom,
\( F_{e}=F_{c} \)
\(\frac{1}{4 \pi \varepsilon_{0}} \frac{e^{2}}{r_{n}^{2}} =\frac{m v_{n}^{2}}{r_{n}} \)
\(\Rightarrow \quad v_{n} =\frac{e}{\sqrt{4 \pi \varepsilon_{0} r_{n} m}} \)      .......(i)
According to the Bohr's second postulate of quantisation,
\(m v_{n} r_{n}=\frac{n h}{2 \pi}\)      .........(ii)
Combining equations (i) and (ii), we get
\( r_{n} =\left(\frac{n^{2}}{m}\right)\left(\frac{h}{2 \pi}\right)^{2} \frac{4 \pi \varepsilon_{0}}{e^{2}} \)
\(=\frac{\varepsilon_{0} n^{2} h^{2}}{\pi m e^{2}} \)      ...........(iii)
\( \because \text { P.E. } =-\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{e^{2}}{r_{n}}=-\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{e^{2}}{\frac{\varepsilon_{0} n^{2} h^{2}}{\pi m e^{2}}} \)
\(=-\frac{m e^{4}}{4 \varepsilon_{0}^{2} n^{2} h^{2}} ; \text { K.E. }=\frac{1}{2} m v_{n}^{2}=\frac{m e^{4}}{8 \varepsilon_{0}^{2} n^{2} h^{2}} \)
Total energy of an electron in the stationary states of hydrogen atom is
\(E_{n}=\text { P.E. }+\mathrm{K} . \mathrm{E} .=\frac{-m e^{4}}{4 \varepsilon_{0}^{2} n^{2} h^{2}}+\frac{m e^{4}}{8 \varepsilon_{0}^{2} n^{2} h^{2}}=\frac{-m e^{4}}{8 \varepsilon_{0}^{2} n^{2} h^{2}} .. ..(iv)\)
According to the third postulate of Bohr's model, \(h v_{i f}=E_{n_{i}}-E_{n_{f}}\)
Using equation above, we get \(h v_{i f}=\frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{2}}\left(\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}}\right)\)
The frequency of radiation emitted is given by \(v_{i f}=\frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{3}}\left(\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}}\right)\)

We have the expression for nuclear radius as:
R = R₀A¹/³
Where,
R₀ = Constant.
A = Mass number of the nucleus
Nuclear matter density, \(\rho \ =\frac{Mass\ of\ the\ nucles}{Volume\ of\ the\ nucles}\)
Let m be the average mass of the nucleus.
Hence, mass of the nucleus = mA
\(\therefore \rho=m \frac{A}{\frac{4}{3} \pi R^{3}}=\frac{3 \mathrm{~mA}}{4 \pi\left(R_{0} A^{\frac{1}{3}}\right)^{3}}=\frac{3 m A}{4 \pi R_{0}^{3} A}=\frac{3 \mathrm{~m}}{4 \pi R_{0}^{3}}\)
Hence, the nuclear matter density is independent of A. It is nearly constant.

we know that velocity of electron moving around a proton in hydrogen atom in an orbit of radius 5.3 × 10^–11 m is 2.2 × 10^–6 m/s. Thus, the frequency of the electron moving around the proton is
\(v=\frac{v}{2 \pi r}=\frac{2.2 \times 10^{6} \mathrm{~m} \mathrm{~s}^{-1}}{2 \pi\left(5.3 \times 10^{-11} \mathrm{~m}\right)}\)
\(\approx \) 6.6 × 10¹⁵ Hz
According to the classical electromagnetic theory we know that the frequency of the electromagnetic waves emitted by the revolving electrons is equal to the frequency of its revolution around the nucleus. Thus the initial frequency of the light emitted is 6.6 × 10¹⁵ Hz.

The angle \(\theta\) made by the area vector of the coil with the magnetic field is 45° . From Eq. (6.1), the initial magnetic flux is
\(\phi\) = BA cos \(\theta\)
\(\begin{aligned} =\frac{0.1 \times 10^{-2}}{\sqrt{2}} \mathrm{Wb} \end{aligned}\)
Final flux, F_min = 0
The change in flux is brought about in 0.70 s. From Eq. (6.3), the magnitude of the induced emf is given by
\(\varepsilon=\frac{\left|\Delta \Phi_B\right|}{\Delta t}=\frac{|(\Phi-0)|}{\Delta t}=\frac{10^{-3}}{\sqrt{2} \times 0.7}=1.0 \mathrm{mV}\)
And the magnitude of the current is
\(I=\frac{e}{R}=\frac{10^{-3} \mathrm{~V}}{0.5 \Omega}=2 \mathrm{~mA}\)
Note that the earth’s magnetic field also produces a flux through the loop. But it is a steady field (which does not change within the time span of the experiment) and hence does not induce any emf.

Capacitance, C = 30μF = 30 × 10⁻⁶F
Inductance, L = 27 mH = 27 × 10⁻³ H
Angular frequency is given as:

\({ \omega }_{ r }=\frac { 1 }{ \sqrt { LC } } =\frac { 1 }{ \sqrt { 27\times { 10 }^{ -3 }\times 30\times { 10 }^{ -6 } } } =\frac { { 10 }^{ 4 } }{ 9 } =1.1\times { 10 }^{ 3 }rad/s\)
Hence, the angular frequency of free oscillations of the circuit is 1.11 × 10³ rad/s.

\(I_{ v }=5.0 \ A, \ v=50 \ c/s, \ I_{ 0 }=\sqrt { 2 } I_{ v }=\sqrt { 2 } \times 5 \ A\)
\(I=I_{ 0 } \ sin \ \omega t= \ I_{ 0 }sin \ 2 \ \pi \ v \ t\)
\(=\sqrt { 2 } \times 5 \ sin \ 2\pi \times 50\times \frac { 1 }{ 300 }\)
\( \\ =5\sqrt { 2 } sin \ \pi /3=5\sqrt { 2 } \times \frac { \sqrt { 3 } }{ 2 } =6.12 \ A\)

Atomic mass of \(\begin{array}{l} 26 \\ 56 \end{array} \text { Fe }\),m₁ = 55.934939 u
\(\begin{array}{l} 26 \\ 56 \end{array} \text { Fe }\) nucleus has 26 protons and (56 − 26) = 30 neutrons
Hence, the mass defect of the nucleus,Δm = 26 x m_H + 30 x m_n − m₁
Where,
Mass of a proton, m_H = 1.007825 u
Mass of a neutron, m_n = 1.008665 u
∴Δm = 26 x 1.007825 + 30 x 1.008665 − 55.934939
= 26.20345 + 30.25995 − 55.934939
= 0.528461 u
But 1 u = 931.5 MeV/c²
∴Δm = 0.528461 x 931.5 MeV/c²
The binding energy of this nucleus is given as:
E_b1 = Δmc²
Where,
c = Speed of light
∴E_b1 = 0.528461 x 931.5 \(\left(\frac{M e V}{c^{2}}\right) \times c^{2}\)
= 492.26 MeV
Average binding energy per nucleon =\(\frac{492.26}{56}=8.79 \mathrm{MeV}\)
Atomic mass of \(\begin{array}{l} 209 \\ 83 \end{array} \text { Bi }\), m₂ = 208.980388 u
\(\begin{array}{l} 209 \\ 83 \end{array} \text { Bi }\) nucleus has 83 protons and (209 − 83) 126 neutrons.
Hence, the mass defect of this nucleus is given as:
Δm' = 83 x m_H + 126 x m_n − m₂
Where,
Mass of a proton, m_H = 1.007825 u
Mass of a neutron, m_n = 1.008665 u
∴Δm' = 83 x 1.007825 + 126 x 1.008665 − 208.980388
= 83.649475 + 127.091790 − 208.980388
= 1.760877 u
But 1 u = 931.5 MeV/c²
∴Δm' = 1.760877 x 931.5 MeV/c²
Hence, the binding energy of this nucleus is given as:
E_b2 = Δm'c²
= 1.760877 x 931.5 \(\left(\frac{M e V}{c^{2}}\right) \times c^{2}\)
= 1640.26 MeV
Average bindingenergy per nucleon \(=\frac{1640.26}{209}=7.848 \mathrm{MeV}\)

For ground level, n₁ = 1
Let E₁ be the energy of this level. It is known that E₁ is related with n₁ as:
\(E_{1}=\frac{-13.6}{n_{1}^{2}} e V\)
\(=-\frac{13.6}{1^{2}}=-13.6 \mathrm{eV}\)
The atom is excited to a higher level, n₂ = 4.
Let E₂ be the energy of this level.
\(\therefore E_{2}=\frac{-13.6}{n_{2}^{2}} e V\)
\(=\frac{-13.6}{4^{2}}=-\frac{13.6}{16} e V\)
The amount of energy absorbed by the photon is given as:
E = E₂ − E₁
\(=-\frac{13.6}{16}-\left(-\frac{13.6}{1}\right)\)
\(=\frac{13.6 \times 15}{16} e V\)
\(=\frac{13.6 \times 15}{16} \times 1.6 \times 10^{-19}=2.04 \times 10^{-18} J\)
For a photon of wavelength λ, the expression of energy is written as:
\(E=\frac{\mathrm{hc}}{\lambda}\)
\(=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{2.04 \times 10^{-18}}\)
= 9.7 x 10^-8 x m= 97nm
And, frequency of a photon is given by the relation,
\(v=\frac{v}{\lambda}\)
\(=\frac{3 \times 10^{8}}{9.7 \times 10^{-8}} \approx 3.1 \times 10^{15} \mathrm{~Hz}\)
Hence, the wavelength of the photon is 97 nm while the frequency is 3.1 x 10¹⁵ Hz.

Given: The values of resistor is 100 Ω and the supply voltage is 100 V.

(a)

The RMS current is given as,
I= V R
Where, the supply voltage is V and the value of resistor is R.
By substituting the given values in the above equation, we get,
I= 220 100 =2.2 A
Thus, the value of RMS current in the conductor is 2.2 A.
(b) Power consumed over a full cycle is given as,
P=VI
Where, the supply voltage is V and the RMS current is I.
By substituting the given values in the above equation, we get
P=220×2.2 =484 W
Thus, power consumed over a full cycle is 484 W.

(i) (a) : From, \(\bar{v}-\frac{1}{\lambda}-R\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)\)
n₁ = 1, n₂ = 2 for first spectral line of Lyman series,
\(\frac{1}{\lambda}=1.097 \times 10^{7}\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)=\frac{3 \times 1.097 \times 10^{7}}{4} \mathrm{~m}^{-1}\)
\(\lambda=\frac{4 \times 10^{-7} \mathrm{~m}}{3 \times 1.097}=\frac{4000}{3 \times 1.097} \dot A=1215.4 \dot A\)
(ii) (d): For wavelength limit, we put \(n_{2}=\infty\)
\(\therefore \quad \frac{1}{\lambda}=1.097 \times 10^{7}\left(\frac{1}{1^{2}}-\frac{1}{\infty}\right)\)
\(\lambda=\frac{1}{1.097 \times 10^{7}} \mathrm{~m}=\frac{1000}{1.097} \dot A=911.6 \dot A\)
(iii) (b): For first line of Balmer series, n₁ = 2, n₂ = 3
\(\bar{v}=\frac{1}{\lambda}=R\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)\)
\(v=\frac{c}{\lambda}=R c\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)\)
\(=1.097 \times 10^{7} \times 3 \times 10^{8}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)\)
\(=1.097 \times 3 \times 10^{15} \times \frac{5}{36}=4.57 \times 10^{14} \mathrm{~Hz}\)
(iv) (d): \(h v_{2 \rightarrow 1}=-13.6\left(\frac{1}{2^{2}}-\frac{1}{1^{2}}\right) \mathrm{eV}=10.2 \mathrm{eV}\)
Emission is n = 2 \(\rightarrow\) n = 1 i.e., higher n to lower n.Transition from lower to higher levels are absorption lines.
\(-13.6\left(\frac{1}{6^{2}}-\frac{1}{2^{2}}\right)=+13.6 \times \frac{2}{9}\)
This is \(
(v) (a): \(\frac{1}{\lambda_{\max }}=R\left[\frac{1}{2^{2}}-\frac{1}{3^{2}}\right]=\frac{5 R}{36}\)
\(\frac{1}{\lambda_{\min }}=R\left[\frac{1}{2^{2}}-\frac{1}{\infty}\right]=\frac{R}{4} \Rightarrow \frac{\lambda_{\min }}{\lambda_{\max }}=\frac{5 R}{36} \times \frac{4}{R}=\frac{5}{9}\)