$$\eqalign{ & \frac{{hC}}{{{\lambda _1}}} - W = \frac{1}{2}mu_1^2 \cr & {\text{and }}\frac{{hC}}{{{\lambda _2}}} - W = \frac{1}{2}mu_2^2 \cr} $$
Dividing the above two equations, we get
$$\eqalign{ & \frac{{\frac{{hC}}{{{\lambda _1}}} - W}}{{\frac{{hC}}{{{\lambda _2}}} - W}} = \frac{{u_1^2}}{{u_2^2}} \cr & \therefore \frac{{\frac{{1240}}{{248}} - W}}{{\frac{{1240}}{{310}} - W}} = \frac{4}{1} \cr & \therefore \frac{{1240}}{{248}} - W = \frac{{4 \times 1240}}{{310}} - 4W \cr & \therefore W = 3.7eV \cr} $$

KEY CONCEPT :
In case of Coolidge tube
$${\lambda _{\min }} = \frac{{hc}}{{eV}} = \lambda \left( {{\text{as}}\,{\text{given}}\,{\text{here}}} \right)$$
Thus the cut off wavelength is inversely proportional to accelerating voltage. As $$V$$ increases, $${\lambda _c}$$ decreases. $${\lambda _k}$$ is the wavelength of $${K_ \propto }$$ line which is a characteristic of an atom and does not depend on accelerating voltage of bombarding electron since $${\lambda _k}$$ always refers to a photon wavelength of transition of $${e^ - }$$ from the target element from $$2 \to 1.$$
The above two facts lead to the conclusion that $${\lambda _k} - {\lambda _c}$$   increases as accelerating voltage is increased.

The continuous spectrum depends on the accelerating voltage. It has a definite minimum wavelength.
Greater the accelerating voltage for electrons, higher will be the kinetic energy it attains before striking the target, higher will be the frequency of X - rays and smaller will be the wavelength. The wavelength of continuous X - rays is independent of the atomic number of target material.

$$p = \frac{E}{c} = \frac{{P \times t}}{c} = \frac{{30 \times {{10}^{ - 3}} \times 100 \times {{10}^{ - 9}}}}{{3 \times {{10}^8}}} = {10^{ - 17}}kg\,m{s^{ - 1}}$$
option (B) is correct

KEY CONCEPT :
$$I = \frac{q}{t} = \frac{{ne}}{t}$$
No. of electrons striking the target per second
$$ = \frac{I}{e} = 2 \times {10^{16}}$$

The order of time is $${10^{ - 10}}s$$

As we know,
$$\left| B \right| = \frac{{\left| E \right|}}{C} = \frac{{6.3}}{{3 \times {{10}^8}}} = 2.1 \times {10^{ - 8}}T$$
As $$\vec V \bot \vec E \bot \vec B$$   therefore direction of is in z direction
$$\vec B = 2.1 \times {10^{ - 8}}\hat kT$$

Initially a photon of energy $$10.2\,eV$$  collides inelastically with a hydrogen atom in ground state. For hydrogen atom,
$$\eqalign{ & {E_1} = - 13.6\,eV;\,{E_2} = - \frac{{13.6}}{4}\,eV = - 3.4\,eV \cr & \therefore {E_2} - {E_1} = 10.2\,eV \cr} $$
The electron of hydrogen atom will jump to second orbit after absorbing the photon of energy $$10.2\,eV.$$  The electron jumps back to its original state in less than microsecond and release a photon of energy $$10.2\,eV.$$  Another photon of energy $$15\,eV$$  strikes the hydrogen atom inelastically. This energy is sufficient to knock out the electron from the atom as ionisation energy is $$13.6\,eV.$$  The remaining energy of $$1.4\,eV$$  is left with electron as its kinetic energy.

Modulated carrier wave contains frequency $${w_c}\,{\text{and}}\,{w_c} \pm {w_m}$$

$$\eqalign{ & \phi = 6.2eV = 6.2 \times 1.6 \times {10^{ - 19}}J \cr & V = 5volt,\frac{{hc}}{\lambda } - \phi = e{V_0} \cr & \Rightarrow \lambda = \frac{{hc}}{{\phi + e{V_0}}} = \frac{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{1.6 \times {{10}^{ - 19}}\left( {6.2 + 5} \right)}} \approx {10^{ - 7}}m \cr} $$