Light of two different frequencies whose photonshave energies 1 e

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 Multiple Choice QuestionsMultiple Choice Questions

131.

Relation between wavelength of photon and electron of same energy is

  • λph > λe

  • λph < λe

  • λph = λe

  • λeλph = constant


132.

Work function of a metal is 5.2 x 10-18, then its threshold wavelength will be

  • 736.7 Ao

  • 760.7 Ao

  • 301 Ao

  • 344.4 Ao


133.

In photoelectric effect, the number of photo-electrons emitted is proportional to

  • velocity of incident beam

  • frequency of incident beam

  • intensity of incident beam

  • work function for cathode material


134.

The graph 1λ and stopping potential (V) of three metals having work function ϕ1 , ϕ2 and  ϕ3 in an experiment of photoelectric effect is plotted as shown in the figure. Which one of the following statement is/are correct? [Here λ is the wavelength of the incident ray ]

     

(i) Ratio of work functions  ϕ1  :  ϕ2  :  ϕ3 = 1 : 2 : 4

(ii) Ratio of work functions ϕ1 :  ϕ2  : ϕ3 = 4 : 2 : 1

(iii) tanθ ∝ h ce, where h = Planck's constant, c = speed of light

(iv)The violet colour-light can eject photoelectrons from metals 2 and 3

  • (i), (iii)

  • (i), (iv)

  • (ii), (iii)

  • (i), (ii) and (iv)


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135.

An isotropic point source emits light with wavelength 500 nm. The radiation power of the source is P = 10 W . Find the number of photons passing through unit area per second at a distance of 3 m from the source.

  • 5.92 x 1017 / m2 s

  • 2.23 x 1017 / m2 s

  • 2.23 x 1018 /m2 s

  • 5.92 x 1018 /m2 s


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136.

Light of two different frequencies whose photons
have energies 1 eV and 2.5 eV respectively, successively illuminate a metallic surface whose work function is 0.5 eV. Ratio of maximum speeds of emitted electrons will be

  • 1 : 4

  • 1 : 1

  • 1 : 5

  • 1 : 2


D.

1 : 2

According to Einstein's photoelectric equation, the maximum kinetic energy of emitted photoelectrons is

         Kmax = h ν  - ϕ0

where hu is the energy of incident photon and ϕ0 is the work function.

    Kmax = hv - ϕ0

∴   12 mvmax2  = hv - ϕ0

As per question

     12 mvmax2 = 1eV - 0.5 eV

                  = 0.5 eV               ....(i)

and 12 mvmax2 = 2.5 eV - 0.5 eV

                  = 2 eV                ......(ii)

Dividing eqn. (i) by eqn. (ii), we get

     vmax12vmax22 = 0.5 eV2 eV

              = 14

   vmax1vmax2 = 14

   vmax1vmax2 = 12


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137.

Assertion:  As work function of a material increases by some mechanism, it requires greater energy to excite the electrons from its surface.

Reason:  A plot of stopping potential (Vs) versus frequency (v) for different metals, has greater slope for metals with greater work functions.

  • If both assertion and reason are true and reason is the correct explanation of assertion.

  • If both assertion and reason are true but reason is not the correct explanation of assertion.

  • If assertion is true but reason is false.

  • If both assertion and reason are false.


138.

Which of the following has the longest de Broglie wavelength if they are moving with the same velocity?

  • Neutron

  • Proton

  • α-particle

  • β-particle


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139.

Assertion: When a certain wavelength of light fall on a metal surface it ejects electron. 

Reason: Light has wave nature.

  • If both assertion and reason are true and reason is the correct explanation of assertion.

  • If both assertion and reason are true but reason is not the correct explanation of assertion.

  • If assertion is true but reason is false.

  • If both assertion and reason are false.


140.

de-Broglie wavelength associated with a ball of mass 1 kg having kinetic energy 0.5 J is

  • 6.63 × 10-30 m

  • 6.63 × 10-34 m

  • 1.32 × 10-33 m

  • 1.32 × 10-34 m


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