1.
A proton has kinetic energy $$E = 100\,keV$$ which is equal to that of a photon. The wavelength of photon is $${\lambda _2}$$ and that of proton is $${\lambda _1}.$$ The ration of $$\frac{{{\lambda _2}}}{{{\lambda _1}}}$$ is proportional to
In a discharge tube, after being accelerated though a high potential difference the ions in the gas strike the cathode with huge kinetic energy. This collision liberates electrons from the cathode. These free electrons can further liberate ions from gas molecules through collisions. The positive ions are attracted towards the cathode and negatively charged electrons move towards anode. Thus, ionisation of gas results.
3.
The wavelength $${\lambda _e}$$ of an electron and $${\lambda _p}$$ of a photon of same energy $$E$$ are related by
A.
$${\lambda _p} \propto \lambda _e^2$$
B.
$${\lambda _p} \propto {\lambda _e}$$
C.
$${\lambda _p} \propto \sqrt {{\lambda _e}} $$
D.
$${\lambda _p} \propto \frac{1}{{\sqrt {{\lambda _e}} }}$$
Wavelength of electron is given by
$${\lambda _e} = \frac{h}{{{p_e}}} = \frac{h}{{\sqrt {2mE} }}\,\,\left[ {\because {p_e} = \sqrt {2mE} } \right]$$
and for photon,
$$\eqalign{
& {\lambda _p} = \frac{{hc}}{E} \Rightarrow \lambda _e^2 = \frac{{{h^2}}}{{2mE}}\,\,{\text{or}}\,\,E = \frac{{hc}}{{{\lambda _p}}} \cr
& \therefore \lambda _e^2 = \frac{{{h^2}}}{{2m \cdot \frac{{hc}}{{{\lambda _p}}}}} \Rightarrow \lambda _e^2 = \frac{{{h^2}}}{{2mhc}}{\lambda _p} \cr
& \Rightarrow \lambda _e^2 \propto {\lambda _p} \cr} $$
4.
An ultraviolet light bulb, emitting $$400\,nm$$ and an infrared light bulb, emitting at $$700\,nm,$$ each are rated at $$130\,W.$$ Then the ratio of the number of photons emitted per second by the $$UV$$ and $$IR$$ sources is -
5.
Find the number of photon emitted per second by a 25 watt source of monochromatic light of wavelength $$6600\,\mathop {\text{A}}\limits^ \circ .$$ What is the photoelectric current assuming $$3\% $$ efficiency for photoelectric effect ?
6.
A particle of mass $$m$$ is projected from ground with velocity making angle $$\theta $$ with the vertical. The de-Broglie wavelength of the particle at the highest point is -
Velocity at highest point $$ = u\sin \theta .$$
$$\therefore {\lambda _D} = \frac{h}{{mu\sin \theta }}\,\,\left( {{\text{Since }}\theta {\text{ is velocity wrt vertical}}} \right)$$
7.
A homogeneous ball (mass = $$m$$) of ideal black material at rest is illuminated with a radiation having a set of photons (wavelength = $$\lambda $$), each with the same momentum and the same energy. The rate at which photons fall on the ball is $$n.$$ The linear acceleration of the ball is
Momentum imparted per unit time $$= np$$
$$\eqalign{
& \Rightarrow F = \frac{{nh}}{\lambda } \cr
& \therefore {\text{Acceleration}} = \frac{{nh}}{{m\lambda }} \cr} $$
8.
When a beam of $$10.6\,eV$$ photons of intensity $$2.0\,W/{m^2}$$ falls on a platinum surface of area $$1.0 \times {10^{ - 4}}{m^2}$$ and work function $$5.6\,eV,0.53\% $$ of the incident photons eject photoelectrons, then the number of photoelectrons emitted per second and their minimum & maximym energies (in $$eV$$ ) [Take $$1\,eV = 1.6 \times {10^{ - 19}}J$$ ] are respectively.
No. of photons/sec $$ = \frac{{{\text{Energy incident on platinum surface per second}}}}{{{\text{Energy of one photon}}}}$$
No. of photons incident per second $$ = \frac{{2 \times 10 \times {{10}^{ - 4}}}}{{10.6 \times 1.6 \times {{10}^{ - 19}}}} = 1.18 \times {10^{14}}$$
As $$0.53\% $$ of incident photon can eject photoelectrons
$$\therefore $$ No. of photoelectrons ejected per second
$$ = 1.18 \times {10^{14}} \times \frac{{0.53}}{{100}} = 6.25 \times {10^{11}}$$
Minimum energy $$= 0\,eV,$$
Maximum energy $$ = \left( {10.6 - 5.6} \right)eV = 5\,eV$$
9.
In a photoemissive cell, with exciting wavelength $$\lambda ,$$ the fastest electron has speed $$v.$$ If the exciting wavelength is changed to $$\frac{{3\lambda }}{4},$$ the speed of the fastest emitted electron will be
A.
$$v{\left( {\frac{3}{4}} \right)^{\frac{1}{2}}}$$
B.
$$v{\left( {\frac{4}{3}} \right)^{\frac{1}{2}}}$$
C.
less than $$v{\left( {\frac{4}{3}} \right)^{\frac{1}{2}}}$$
D.
greater than $$v{\left( {\frac{4}{3}} \right)^{\frac{1}{2}}}$$
Answer :
greater than $$v{\left( {\frac{4}{3}} \right)^{\frac{1}{2}}}$$
Einstein's photoelectric equation is given by
$$KE = E - {W_0}$$
As we know that
$$\eqalign{
& KE = \frac{1}{2}m{v^2}\,\,{\text{and}}\,\,E = \frac{{hc}}{\lambda } \cr
& \therefore \frac{1}{2}m{v^2} = \frac{{hc}}{\lambda } - {W_0}......\left( {\text{i}} \right) \cr} $$
Suppose $${v'}$$ be the new speed, when $$\lambda $$ is changed to $$\frac{{3\lambda }}{4}.$$
The new equation can be written as
$$\frac{1}{2}m{{v'}^2} = \frac{{hc}}{{\left( {\frac{{3\lambda }}{4}} \right)}} - {W_0}$$
$${\text{or}}\,\,\frac{1}{2}m{{v'}^2} = \frac{4}{3}\frac{{hc}}{\lambda } - {W_0}\,......\left( {{\text{ii}}} \right)$$
Dividing Eq. (ii) by Eq. (i), we get
$$\eqalign{
& \frac{{{{v'}^2}}}{{{v^2}}} = \frac{{\frac{4}{3}\frac{{hc}}{\lambda } - {W_0}}}{{\frac{{hc}}{\lambda } - {W_0}}} \cr
& = \frac{{\frac{4}{3}\frac{{hc}}{\lambda } - \frac{4}{3}{W_0} + \frac{1}{3}{W_0}}}{{\frac{{hc}}{\lambda } - {W_0}}} \cr
& = \frac{4}{3} + \frac{{{W_0}}}{{3\left( {\frac{{hc}}{\lambda } - {W_0}} \right)}} > \frac{4}{3} \cr
& \therefore \frac{{v'}}{v} > \sqrt {\frac{4}{3}} \,\,{\text{or}}\,\,v' > \sqrt {\frac{4}{3}v} \cr} $$
10.
Doubly ionised helium atoms and hydrogen ions are accelerated from rest through the same potential drop. The ratio of the final velocities of the helium and the hydrogen ion is
If $$v$$ is the speed acquired by particle when accelerated under a potential difference $$V,$$ then
$$KE$$ gained by particle $$ = \frac{1}{2}m{v^2} = eV$$
or $$v = \sqrt {\frac{{2\,eV}}{m}} $$
So, for two different cases of the He-atom and H-atom,
$$\frac{{{v_{He}}}}{{{v_H}}} = \frac{{\sqrt {\frac{{2{{\left( e \right)}_{He}}V}}{{{m_{He}}}}} }}{{\sqrt {\frac{{2{{\left( e \right)}_H}V}}{{{m_H}}}} }}$$
As $${m_{He}} = 4\,{m_H}\,\,{\text{and}}\,\,{\left( e \right)_{He}} = 2{\left( e \right)_H}$$
$$\therefore \frac{{{v_{He}}}}{{{v_H}}} = \frac{1}{{\sqrt 2 }}$$