The minimum wavelength of $X$-rays produced in a coolidge tube operated at potential difference of $40\;kV$ is

${\lambda _{{\rm{min}}}} = \frac{{12375}}{{40 \times {{10}^3}}} = 0.309\;\mathop A\limits^ \circ \approx 0.31\;\mathop A\limits^ \circ $

The correct Option is "$0.31\;\mathop A\limits^ \circ $"

$3.1\;\mathop A\limits^ \circ $

$311\;\mathop A\limits^ \circ $

In an $X$-rays tube, the intensity of the emitted $X$-rays beam is increased by

By changing the filament current with the help of rheostat, thermionic emission intensity of $X$-rays can be changed

The correct Option is "Increasing the filament current"

Decreasing the filament current

Increasing the target potential

Decreasing the target potential

In Davisson - Germer experiment maximum intensity is observed at

We know that in Davisson-Germer experiment maximum intensity is observed at ${54^{^ \circ }}$ and 50 V.

The correct Option is "${54^ \circ }$ and 50 V"

For characteristic $X$-ray of some material

${ \because }E\left( {{K_\gamma }} \right) > E\left( {{K_\beta }} \right) > E\left( {{K_\alpha }} \right) \Rightarrow \lambda \left( {{K_\gamma }} \right) < \lambda \left( {{K_\beta }} \right) < \lambda \left( {{K_\alpha }} \right)$

$E\left( {{K_\gamma }} \right) < E\left( {{K_\beta }} \right) < E\left( {{K_\alpha }} \right)$

$E\left( {{K_\alpha }} \right) < E\left( {{L_\alpha }} \right) < E\left( {{M_\alpha }} \right)$

The correct Option is "$\lambda \left( {{K_\gamma }} \right) < \lambda \left( {{K_\beta }} \right) < \lambda \left( {{K_\alpha }} \right)$"

$\lambda \left( {{M_\alpha }} \right) < \lambda \left( {{L_\alpha }} \right) < \lambda \left( {{K_\alpha }} \right)$

The continuous X-rays spectrum produced by an X-ray machine at constant voltage has

Continuous spectrum of $X$-rays consists of radiations of all possible wavelength range having a definite short wavelength limit

The correct Option is "A minimum wavelength"

The intensity of X-rays from a coolidge tube is plotted against wavelength $\lambda $ as shown in figure. The minimum wavelength found is ${\lambda _c}$ and the wavelength of ${K_{\rm{\alpha }}}$ like is ${\lambda _K}$. As the accelerating voltage is increased

As accelerating voltage $V$ across $X$-rays tube increase, the value of minimum wavelength of $X$-rays, ${\lambda _c} = \frac{{hc}}{{eV}};$ decreases ; so the separation between ${\lambda _K}$ and ${\lambda _c}$ increases.

The correct Option is "${\lambda _K} - {\lambda _c}$ increases"

${\lambda _K} - {\lambda _c}$ decreases

An X-ray tube with a copper target emits ${K_\alpha }$ line of wavelength $1.50\;\mathop A\limits^ \circ .$ What should be the minimum voltage through which electrons are to be accelerated to produce this wavelength of ${\rm{X}}$ rays $(h = 6.63 \times {10^{ - 34}}J$-$s,\;c = 3 \times {10^8}m/s)$

$eV = \frac{{hc}}{\lambda } = \frac{{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{1.5 \times {{10}^{ - 10}}}}$ $ \Rightarrow V = \frac{{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{1.6 \times {{10}^{ - 19}} \times 1.5 \times {{10}^{ - 10}}}} = 8280\;Volt$

The correct Option is "$8280\;V$"

In an ionisation experiment it is found that a doubly ionised particle enters a magnetic field of 1 T and moves in a circular path of radius 1 m with a speed of $1.6 \times {10^7}\;{\rm{m}}{{\rm{s}}^{ - 1}}$. The particle must be

As $q\;v\;B = m{v^2}/r$ or $m = \frac{{qBr}}{v} = \frac{{\left( {2 \times 1.6 \times {{10}^{ - 19}}} \right) \times 1 \times 1}}{{1.6 \times {{10}^7}}}$ $ = 2 \times {10^{ - 26}}{\rm{kg}} = \frac{{2 \times {{10}^{26}}}}{{1.66 \times {{10}^{ - 27}}}} = 12$ Therefore, particle must be ${{\rm{c}}^{ + + }}$

The correct Option is "${{\rm{C}}^{ + + }}$"

An X-ray machine is operated at 40 kV. The short wavelength limit of continuous X-rays will be $\left( {h = 6.63 \times {{10}^{ - 34}}Js,\;c\; = \;3 \times {{10}^8}\;m{s^{ - 1}},\;e = 1.6 \times {{10}^{ - 19}}\;C} \right)$

For an accelerating voltage $V,\;$ the maximum X-ray photon energy is given by $\;hv{\;_{{\rm{max}}}} = eV$ Also, $\gamma = $ $\frac{c}{\lambda } = \frac{{{\rm{velocity\;of\;light\;}}}}{{{\rm{wavelength\;}}}}$ $\therefore \;\;\;\;\;\;{\rm{\lambda }}{{\rm{\;}}_{{\rm{min}}}} = \frac{{{\rm{ch}}}}{{{\rm{eV}}}} = \frac{{3 \times {{10}^8} \times 6.63 \times {{10}^{ - 34}}}}{{1.6 \times {{10}^{ - 19}} \times 40 \times {{10}^3}}}$ = $0.31 \times {10^{ - 10}}$ m $ \approx 0.31$ A

The correct Option is "$0.31\mathop A\limits^ \circ $"

$0.155\mathop A\limits^ \circ $

Penetrating power of X-rays can be increased by

With the increase in potential difference between anode and cathode energy of striking electrons increases which in turn increses the energy (penetration power) of $X$-rays