Q: The given graph shows the variation of photo-electric current \left(I\right) versus applied voltage \left(V\right) for two different photosensitive materials and for two different intensities of the incident radiation. Identify the pairs of curves that correspond to different materials but same intensity of incident radiation.

Since curve 1 and curve 2 have the same cutoff value, they belong to the same material. Similarly, curve 3 and curve 4 also have the same cutoff voltage, so they also belong to the same material. At V=0, the curves 1 and 3, and 2 and 4 move together, therefore, curves 1 and 3 and curves 2 and 4 correspond to different materials but same intensity of incident radiation.

Q: Davisson and Germer studied the diffraction from crystal of beams of

Davisson and Germer's actual objective was to study the surface of a piece of nickel by directing a beam of electrons at the surface and observing how many electrons bounced off at various angles. They expected that because of the small size of electrons, even the smoothest crystal surface would be too rough and thus the electron beam would experience diffuse reflection.During the experiment an accident occurred and air entered the chamber, producing an oxide film on the nickel surface. To remove the oxide, Davisson and Germer heated the specimen in a high temperature oven, not knowing that this affected the formerly polycrystalline structure of the nickel to form large single crystal areas with crystal planes continuous over the width of the electron beam. When they started the experiment again and the electrons hit the surface, they were scattered by atoms which originated from crystal planes inside the nickel crystal.In 1925, they generated a diffraction pattern with unexpected peaks.So, the answer is option (C).

Q: In Davisson and Germer experiment first order diffraction is observed. Accelerating voltage is 54 volt. If the distance between reflecting planes of Ni crystal is 0.92 \mathring A then find out diffraction angle.

Wavelength of electrons (\lambda) = \dfrac{12.27}{\sqrt{V}} \mathring A = \dfrac{1227}{\sqrt{54}} \mathring A = \dfrac{12.27}{7.348} \mathring A = 1.669 \mathring A \approx 1.67 \mathring A Perpendicular distance in diffraction pattern = d \, sin \theta Perpendicular distance for first order = \dfrac{\lambda}{2} \therefore sin \theta = \dfrac{\lambda}{2d} = \dfrac{1.67 \mathring A}{2 \times 0.92 \mathring A} = 0.907 \theta = 65^{\circ}

Q: The mass of a photon in motion is (given its frequency = x ):

By Einstein's mass energy equivalence E=mc^{2} but E=hx \therefore hx=mc^{2} \Rightarrow m=\dfrac{hx}{c^{2}}

Q: Light of wavelength 2200A^o (angstorm) falls on a photosensitive plate with work function 4.1 eV. Find (a) energy of photon in eV (electron volt). (b) maximum kinetic energy of photoelectron and (c) stopping potential.

Energy of photon= \frac{hc}{\lambda}joules \lambda=2200A^{0} Work function \phi=4.1eV 1eV=1.6\times 10^{-19} joules (a) E = \dfrac{12375}{\lambda(A^\circ )}eV E=5.6eV (b)maximum Kinetic energy KE_{max}=E-\phi KE_{max}=5.6-4.1=1.5eV (c)stopping potential= \dfrac{KE_{max}}{e} stopping potential =1.5V

Q: Find the number of photon emitted per second by a 25\ watt source of monochromatic light of wavelength 6600\overset {o}{A} . What is the photoelectric current assuming 3\% efficiency for photoelectric effect?

P_{in} = 25W, \lambda = 6600\overset {\circ}{A} = 6600\times 10^{-10} m nhv = P \Rightarrow Number of photons emitted / sec, n = \dfrac {P}{\dfrac {hc}{\lambda}} = \dfrac {P\lambda}{hc} = \dfrac {25\times 6600\times 10^{-10}}{6.64\times 10^{-34} \times 3\times 10^{8}} = 8.28\times 10^{19} = \dfrac {25}{3} \times 10^{19} 3 % of emitted photons are producing current \therefore I = \dfrac {3}{100} \times ne = \dfrac {3}{100}\times \dfrac {25}{3} \times 10^{19}\times 1.6\times 10^{-19} = 0.4\ A .

Q: Choose the correct answer from the alternatives given.A parallel beam of monochromatic radiation of cross-section area A ( \, \phi_0) and radius a. The sphere is grounded by a conducting wire. Assume that for each incident photon, one photoelectron is ejected. Just after this radiation is incident on initially unchanged sphere the current through the conducting wire is

Energy of each incident photon = h\upsilon Total energy incident per second on sphere = IA \therefore No. of photons incident on sphere per second = \dfrac{IA}{h\upsilon} \therefore No. of photolectrons ejected per second = \dfrac{IA}{h\upsilon} Current through the conducting wire = charge flowing per second through wire = \dfrac{IAe}{h\upsilon}

Question 1

A graph is plotted between uncertainty in position and inverse of uncertainty in wavelength for an electron. We get a straight line passing through the origin. Calculate voltage through which electron is accelerated with

Accepted Answer

Slope = \displaystyle\frac { \Delta x }{ { 1 }/{ \Delta \lambda  } } =\Delta x\cdot \Delta \lambda  By Heisenberg principle  \Delta x\cdot \Delta \lambda = \displaystyle\frac { { \lambda  }^{ 2 } }{ 4\pi  }  \displaystyle\frac { { \lambda  }^{ 2 } }{ 4\pi  } = \displaystyle\frac { 1 }{ \pi  }                 \Rightarrow   \lambda =2\mathring { A }  \lambda = \sqrt { \displaystyle\frac { 150 }{ V }  }              \Rightarrow V=37.5V

Question 2

The figure shows graph between stopping potential (along  y -axis) and incident frequency (along x-axis) for different target metal  A, B, C  and  D . Let  \displaystyle \phi _{A},\phi _{B},\phi _{C}\: and\: \phi _{D}  are the work function of different target metals  A, B, C  and  D  respectively Then the correct rank of work function of different target metals is

Accepted Answer

The Einstein's photoelectric equation in terms of stopping potential is  eV_s=h
u-\phi   or  V_s=\dfrac{h
u}{e}-\phi/e Thus, it will represent a straight line and intercept is  \phi/e From the graph we can write,  \phi_A/e>\phi_B/e>\phi_C/e>\phi_D/e or  \phi_A>\phi_B>\phi_C>\phi_D

Question 3

Which one of the following graph represents the variation of maximum kinetic energy  { (E }_{ K })  of the emitted electrons with frequency  v  in photoelectric effect correctly?

Accepted Answer

Using Einstein's Law of Photo-electric effect: \Rightarrow h
u=W+E_k  where  W  is the work function and  h  is the Plank's constant.From the above equation, the curve  E_k  vs  
u  is a straight line with slope  h .Here we can see that  E_k  will be negative for  h
u<W  which we do not consider.Graph starts at the point  (\dfrac{W}{h},0)  and is straight line with slope  h .

Question 4

The given graph shows the variation of photo-electric current  \left(I\right)  versus applied voltage  \left(V\right)  for two different photosensitive materials and for two different intensities of the incident radiation. Identify the pairs of curves that correspond to different materials but same intensity of incident radiation.

Accepted Answer

Since curve 1 and curve 2 have the same cutoff value, they belong to the same material. Similarly, curve 3 and curve 4 also have the same cutoff voltage, so they also belong to the same material. At V=0, the curves 1 and 3, and 2 and 4 move together, therefore, curves 1 and 3 and curves 2 and 4 correspond to different materials but same intensity of incident radiation.

Question 5

Davisson and Germer studied the diffraction from crystal of beams of

Accepted Answer

Davisson and Germer's actual objective was to study the surface of a piece of nickel by directing a beam of electrons at the surface and observing how many electrons bounced off at various angles. They expected that because of the small size of electrons, even the smoothest crystal surface would be too rough and thus the electron beam would experience diffuse reflection.During the experiment an accident occurred and air entered the chamber, producing an oxide film on the nickel surface. To remove the oxide, Davisson and Germer heated the specimen in a high temperature oven, not knowing that this affected the formerly polycrystalline structure of the nickel to form large single crystal areas with crystal planes continuous over the width of the electron beam. When they started the experiment again and the electrons hit the surface, they were scattered by atoms which originated from crystal planes inside the nickel crystal.In 1925, they generated a diffraction pattern with unexpected peaks.So, the answer is option (C).

Question 6

In Davisson and Germer experiment first order diffraction is observed. Accelerating voltage is  54  volt. If the distance between reflecting planes of Ni crystal is  0.92 \mathring A  then find out diffraction angle.

Accepted Answer

Wavelength of electrons   (\lambda) = \dfrac{12.27}{\sqrt{V}} \mathring A      = \dfrac{1227}{\sqrt{54}} \mathring A = \dfrac{12.27}{7.348} \mathring A = 1.669 \mathring A        \approx 1.67 \mathring A   Perpendicular distance in diffraction pattern         = d \, sin 	heta   Perpendicular distance for first order   = \dfrac{\lambda}{2}     	herefore                 sin 	heta = \dfrac{\lambda}{2d} = \dfrac{1.67 \mathring A}{2 	imes 0.92 \mathring A} = 0.907                                  	heta = 65^{\circ}

Question 7

The mass of a photon in motion is (given its frequency  =  x ):

Accepted Answer

By Einstein's mass energy equivalence  E=mc^{2} but  E=hx 	herefore hx=mc^{2} \Rightarrow m=\dfrac{hx}{c^{2}}

Question 8

Light of wavelength  2200A^o  (angstorm) falls on a photosensitive plate with work function  4.1  eV. Find (a) energy of photon in eV (electron volt). (b) maximum kinetic energy of photoelectron and (c) stopping potential.

Accepted Answer

Energy of photon= \frac{hc}{\lambda}joules \lambda=2200A^{0} Work function  \phi=4.1eV 1eV=1.6	imes 10^{-19} joules (a)  E = \dfrac{12375}{\lambda(A^\circ )}eV E=5.6eV (b)maximum Kinetic energy  KE_{max}=E-\phi KE_{max}=5.6-4.1=1.5eV (c)stopping potential= \dfrac{KE_{max}}{e} stopping potential =1.5V

Question 9

Find the number of photon emitted per second by a  25\ watt   source of monochromatic light of wavelength  6600\overset {o}{A} . What is the photoelectric current assuming  3\%  efficiency for photoelectric effect?

Accepted Answer

P_{in} = 25W, \lambda = 6600\overset {\circ}{A} = 6600	imes 10^{-10} m nhv = P \Rightarrow  Number of photons emitted / sec, n = \dfrac {P}{\dfrac {hc}{\lambda}} = \dfrac {P\lambda}{hc} = \dfrac {25	imes 6600	imes 10^{-10}}{6.64	imes 10^{-34} 	imes 3	imes 10^{8}} = 8.28	imes 10^{19} = \dfrac {25}{3} 	imes 10^{19} 3 % of emitted photons are producing current 	herefore I = \dfrac {3}{100} 	imes ne = \dfrac {3}{100}	imes \dfrac {25}{3} 	imes 10^{19}	imes 1.6	imes 10^{-19} = 0.4\ A .

Question 10

Choose the correct answer from the alternatives given.A parallel beam of monochromatic radiation of cross-section area A  ( \, \phi_0)  and  radius a. The sphere is grounded by a conducting wire. Assume that for each incident photon, one photoelectron is ejected. Just after this radiation is incident on initially unchanged sphere the current through the conducting wire is

Accepted Answer

Energy of each incident photon =  h\upsilon Total energy incident per second on sphere = IA 	herefore  No. of photons incident on sphere per second =  \dfrac{IA}{h\upsilon} 	herefore  No. of photolectrons ejected per second =  \dfrac{IA}{h\upsilon} Current through the conducting wire  = charge flowing per second through wire =  \dfrac{IAe}{h\upsilon}

Question 11

Select the right one from the given options.

Accepted Answer

Young's Double Slit experiment is a conclusive proof supporting wave nature of light.

Question 12

If we assume that penetrating power of any radiation/particle is inversely proportional to the De-broglie wavelength of the particle, then

Accepted Answer

\lambda = \dfrac{h}{P} = \dfrac{h}{\sqrt{2mE}} \Rightarrow \dfrac{h}{\sqrt{2m(Vq)}} 	herefore  For higher m and q ; \lambda   will be smaller.For an ' \alpha ' particle; both 'm' and 'q' are higherhence lesser is the wavelength.As, (penetrating power)  \propto  Energy  \propto \dfrac{1}{\lambda} From above; penetrating power of an  \alpha- particle is more than that of a proton.So, the answer is option (B).

Question 13

De Broglie proposed the wave nature of electrons suggesting matter waves.Find the momentum, speed and De-Broglie wavelength of an electron with Kinetic energy of  120 eV .

Accepted Answer

Here ,   E=120eV=120	imes1.6	imes10^{-19}J  ,  m_{e}=9.1	imes10^{-31}kg  ,   h=6,6	imes10^{-34}Js Now , if  v  be the speed of electron , then             E=1/2m_{e}v^{2} or          v=\sqrt{2E/m_{e}} or          v=\sqrt{2	imes120	imes1.6	imes10^{-19}/9.1	imes10^{-31}} or          v=6.49	imes10^{6}m/s  Now , momentum of electron,              p=m_{e}v=9.1	imes10^{-31}	imes6.49	imes10^{6}=5.9	imes10^{-24}kgm/s de-Broglie wavelength of the electron,              \lambda=h/p=6.6	imes10^{-34}/5.9	imes10^{-24}=1.1	imes10^{-10}m

Dual Nature Of Radiation And Matter