X-rays: They are generated by bombarding a target of high atomic number Z with a beam of fast moving electrons. Uses : (i) radio theapy for curing skin diseases. (ii) Medical diagnosis locating fracture TB etc.

(i) The basic difference between various type of electromagnetic waves lies in their wavelength of frequencies, since all of them travel through vacuum with the same speed. Consequently, the waves differ considerably in their mode of interaction with matter.
(ii) If there is no water present in the food material, then microwaves will not be able to generate heat in the food.
(iii) It is very necessary to wear special glass googles to welders while working, so that they can protect their eyes from harmful electromagnetic radiation.

(a) \(\gamma\)-rays. Use: In treatment of cancer.
(b) Microwaves. Use: In radar system for aircraft navigation.
(c) X-rays. Use: As a diagnostic tool for the detection of fractures.

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.

Total energy of the electron in hydrogen atom is –13.6 eV = –13.6 × 1.6 × 10^–19 J = –2.2 ×10^–18 J. Thus from Equation we have
\(-\frac{e^{2}}{8 \pi \varepsilon_{0} r}=-2.2 \times 10^{-18} \mathrm{~J}\)
This gives the orbital radius
\(r=-\frac{e^{2}}{8 \pi \varepsilon_{0} E}=-\frac{\left(9 \times 10^{9} \mathrm{~N} \mathrm{~m}^{2} / \mathrm{C}^{2}\right)\left(1.6 \times 10^{-19} \mathrm{C}\right)^{2}}{(2)\left(-2.2 \times 10^{-18} \mathrm{~J}\right)}\)
= 5.3 × 10^–11 m
The velocity of the revolving electron can be computed from Eq.uation with m = 9.1 × 10^–31 kg,
\(v=\frac{e}{\sqrt{4 \pi \varepsilon_{0} m r}}=2.2 \times 10^{6} \mathrm{~m} / \mathrm{s}\)

In α-decay, the atomic number is decreased by 2 units and mass number decreases by 4 units. In β-decay, the atomic number increases by I unit but mass number does not change. In ⋎-decay, there is no change in mass and atomic number.
Therefore, the mentioned radioactive decays will proceed as below.
\(_{ 71 }^{ 176 }{ { A }_{ 2 } }+_{ -1 }^{ 0 }{ \beta }\longrightarrow _{ 70 }^{ 176 }{ { A }_{ 1 } }+_{ 2 }^{ 4 }{ He }\longrightarrow _{ 72 }^{ 180 }{ { A } }\)
So, A has mass number 180 and atomic number 72.
Also 
\(_{ 71 }^{ 176 }{ { A }_{ 2 } }\overset { \alpha }{ \longrightarrow } \)\(_{ 69 }^{ 172 }{ { A }_{ 3 } }+\overset { \gamma }{ \longrightarrow } \)\(_{ 69 }^{ 172 }{ { A }_{ 4 } }\)
So, A₄ has mass number 172 and atomic number 69.
(ii) ß^-decay
\(\underset { (neutron)\\ decay }{ n } \rightarrow \underset { (proton) }{ p } +{ \beta }^{ - }+{ v }^{ - }\) (anti-neutrino)
e.g. \(_{ 15 }^{ 32 }{ P }\rightarrow _{ 16 }^{ 32 }{ S+{ e }^{ - }+\overset { - }{ v } }\)
ß⁺ - decay
\(\underset { (proton \ decay) }{ p } \rightarrow \underset { (neutron) }{ n } +{ \beta }^{ + }+{ v }(neutrino)\)
e.g \(_{ 11 }^{ 22 }{ N }\rightarrow _{ 10 }^{ 22 }{ Ne+{ e }^{ + }+v }\)

Number of half lives of P in 18 hours
Number of half lives of Q in 18 hours
Therefore, no. of nuclei left undecayed:
\({ N }_{ 1 }={ N }_{ 0 }{ \left( \frac { 1 }{ 2 } \right) }^{ 6 }={ N }_{ 0 }/64\)
\( { N }_{ 2 }={ N }_{ 0 }{ \left( \frac { 1 }{ 2 } \right) }^{ 2 }={ N }_{ 0 }/64\)
The ratio of their rates of disintegration is
\( \frac { { R }_{ 1 } }{ { R }_{ 2 } } =\frac { { \lambda }_{ 1 }{ N }_{ 1 } }{ { \lambda }_{ 2 }{ N }_{ 2 } } =\frac { { T }_{ 2 } }{ { T }_{ 1 } } \left( \frac { { N }_{ 1 } }{ { N }_{ 2 } } \right) =\frac { 9 }{ 3 } \times \frac { { N }_{ 0 }/64 }{ { N }_{ 0 }/4 } =\frac { 3 }{ 16 } \)

\(36\times { 10 }^{ 23 },36\times { 10 }^{ 23 },48\times { 10 }^{ 23 }\)
Each atom of  \(_{ 6 }{ { C }^{ 14 } }\) , 8 neutrons, and 6 electrons and there are \(=6\times { 10 }^{ 23 }\) such atoms in 14 gm of Carbon 14.

\(4.4\times { 10 }^{ -36 }\ kg\)
\(E=hv=m{ c }^{ 2 }\)
\(\therefore \ m=\frac { hv }{ { c }^{ 2 } } =\frac { 6.6\times { 10 }^{ -34 }\times 6\times { 10 }^{ 14 } }{ { \left( 3\times { 10 }^{ 8 } \right) }^{ 2 } } \)
\(=4.4\times { 10 }^{ -36 } \ kg\)

\(2.11\times { 10 }^{ -34 }Js\)
\(E=\frac { -13.6 }{ { n }^{ 2 } } =-3.4eV \ \therefore n=2\)
\(\therefore \) Angular momentum of electron 
\(=\frac { nh }{ 2\pi } ={ 2h }/{ 2\pi }={ h }/{ \pi }=\frac { 6.626\times { 10 }^{ -34 } }{ 3.14 }\)
\( \\ =2.11\times { 10 }^{ -34 }J^{-s}\)

Using Eq, the magnitude of B is
\(B=\frac { E }{ c } \) 
\(=\frac { 6.3V/m }{ 3\times { 10 }^{ 8 }m/s } =2.1\times { 10 }^{ -8 }T\)
To find the direction, we note that E is along y-direction and the wave propagates along x-axis. Therefore, B should be in a direction perpendicular to both x- and y-axes. Using vector algebra, E × B should be along x-direction.
Since, \((+\overrightarrow{\mathbf{j}}) \times(+\hat{\mathbf{k}})=\overrightarrow{\mathbf{i}}, \mathbf{B}\) is along the z-direction.
Thus, \(\mathbf{B}=2.1 \times 10^{-8} \hat{\mathbf{k}} \mathrm{T}\)

\(3646.8\mathring { A } \)
\(Take \ { n }_{ 1 }=2 \ and \ { n }_{ 2 }=\infty \)

\(7.7\times { 10 }^{ -40 }{ m }^{ 3 }\)
Maximum volume in which positive charge of atom is likely to be concentrated is
\({ V }_{ 0 }=\frac { 4 }{ 3 } \pi { r }_{ 0 }^{ 3 }\)
Calculate \({ r }_{ 0 }\) from the usual formula.

5.688 MeV

According to Bohr's quantum condition, the permitted orbits are those in which the angular momentum of the electron is integral multiple of \(nh/2\pi\), where h is Planck's constant
i.e. m v r = \(nh/2\pi\) (when n = 1, 1, ............ called principal quantum number).

If r₀ is Bohr radius and v₀ is the velocity of electron in 1st orbit, then
Time period \(T={2\pi r_0\over v_0}\)
So   \(I={e\over T}={ev_0\over2\pi r_0}\)

It is given that a plane electromagnetic wave travels in vacuum along z-direction and the frequency of the electromagnetic wave is 30MHz.
We can say that electric field and magnetic field will be in x-plane because the electromagnetic wave travels along the z-direction and both fields are mutually perpendicular to each other.
The formula of the wavelength of a wave is,
λ=c/ν
Substitute the values in the above expression,
λ=(3×10⁸)/(30×10⁶)=10m.
Thus, the value of wavelength is 10 m and the direction of electric and magnetic fields will be in x-y plane.

The total energy falling on the surface is
\(U=(18W/{ cm }^{ 2 })\times (20{ cm }^{ 2 })\times (30\times 60)\)
\( =6.48\times 10^{ 5 }J\)
Total momentum delivered (for complete absorption) is
\(P=\frac { U }{ c } =\frac { 6.48\times { 10 }^{ 5 }J }{ 3\times { 10 }^{ 8 }m/s } =2.16\times { 10 }^{ -3 } \ kgm/s\) 
The average force exerted on the surface is
\(F=\frac { P }{ t } =\frac { 2.16\times 10^{ -3 } }{ 0.18\times { 10 }^{ 4 } } =1.2\times 10^{ -6 } \ N\) 
If the surface is perfect reflector, the change in momentum 
\(=p-(-p)=2p=2+2.16\times { 10 }^{ -3 } \ kg \ ms^{ -1 }\) 
And average force
 \(F=\frac { 2p }{ t } =\frac { 2\times 2.16\times { 10 }^{ -3 } }{ 30\times 60 } \)
\(=2.4\times 10^{ -6 }N\)

Given \(B=8\times 10^{ 6 } \ sin \ (2\times 10^{ 11 }t+300\pi x) \ T\) wave from equation of magnetic field.
\({ B }_{ y }={ B }_{ 0 } \ sin \ \left( \frac { 2\pi }{ T } t+\frac { 2\pi }{ \lambda } x \right) \) 
or  \(300\pi =\frac { 2\pi }{ \lambda } \) 
or  \(\lambda =\frac { 2\pi }{ 300\pi } =\frac { 1 }{ 150 } =6.3\times { 10 }^{ -3 } \ m\) 
(ii) Electric field is given as
\({ E }_{ Z }={ B }_{ Z } \ c=24\times 10^{ 14 }sin\left[ 2\times 10^{ 11 }t+300\pi x \right] Vm^{ -1 }\)

\({ I }_{ av }=\frac { c }{ 2 } \frac { { B }^{ 2 }_{ 0 } }{ { \mu ^{ 2 } }_{ 0 } } =\frac { 3\times 10^{ 8 }\times (12\times 10^{ 8 })^{ 2 } }{ 2\times 4\pi \times 10^{ -7 } } =1.7Wm^{ -2 }\)