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Dual Nature of Matter and Radiation Test - 11

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Dual Nature of Matter and Radiation Test - 11
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  • Question 1
    1 / -0.25

    In a photo-emissive cell, with exciting wavelength l, the maximum kinetic energy of electron is K. If the exciting wavelength is changed to  the kinetic energy of the fastest emitted electron will be :

    Solution

    From E=W0 ​ +(1/2)​mvmax2 ​⇒vmax ​= √[(2E/m)​−(2W0 /m)]​​​
    where E= hc ​/λ
    If wavelength of incident light charges from λto 3 λ/4 ​(Decreases)
    Let energy of incident light charges from E and speed of fastest electron changes from v to v ′then
    v ′=√[(2E ′/m)​−(2W0 /m)​​​]
    As E ∝1/λ​⇒E ′(4/3)​E
    Hence v ′=√[(2(4/3 ​)E/m)​−(2W0 /m)​​]​
    ⇒v ′=(4/3)1/2 [(2E/m)​−(2W0 /m(4/3)1/2 ​)]​​
    So, v ′>(4/3)1/2 v

  • Question 2
    1 / -0.25

    The frequency and the intensity of a beam of light falling on the surface of photoelectric material are increased by a factor of two (Treating efficiency of photoelectron generation as constant). This will :

    Solution

  • Question 3
    1 / -0.25

    Light coming from a discharge tube filled with hydrogen falls on the cathode of the photoelectric cell. The work function of the surface of cathode is 4eV. Which one of the following values of the anode voltage (in Volts) with respect to the cathode will likely to make the photo current zero.

    Solution

    Given, 
    Work function, ϕ=4eV
    for Hydrogen atom, it can provide a max of 13.6 eV of energy by exciting its electron.
    ∴e(Stopping potential) ≥13.6 - 4
    ≥9.6
    i.e. V = 10 V will make photo current zero.

  • Question 4
    1 / -0.25

    Let K1  be the maximum kinetic energy of photoelectrons emitted by a light of wavelength l1  and K2  corresponding to l2 . If l1  = 2l2 , then :

    Solution

    We know that,
    [(hc/ λ) -Wo ]=K1 =>l1 =hc/(K1 +Wo )
    [(hc/l2 )-Wo]=K2 =>l2 = hc/(K2+ Wo )
    Given, l1 =2l2
    =>hc/K2 + Wo =2hc/(K2 + Wo )
    Or,K1 +Wo= (K2 + Wo )/2
    Or,K1 =(K2 /2)-(Wo/2 )
    Since, Wo >1=>K1 2 /2

  • Question 5
    1 / -0.25

     

    In a photoelectric experiment, the potential difference V that must be maintained between the illuminated surface and the collector so as just to prevent any electron from reaching the collector is determined for different frequencies f of the incident illumination. The graph obtained is shown. The maximum kinetic energy of the electrons emitted at frequency f1  is

        

     

    Solution

     

     

     

     

  • Question 6
    1 / -0.25

     

    In photoelectric effect, stopping potential depends on

     

    Solution

     

     

    The stopping potential depends on frequency of incident light and nature of the emitter material.for a given frequency of incident light, it is independent of its intensity.

     

     

  • Question 7
    1 / -0.25

     

     In the experiment on photoelectric effect using light having frequency greater than the threshold frequency, the photocurrent will certainly increase when

     

    Solution

     

     

    If light of certain frequency, greater than the threshold frequency, is incident of the metal surface the photoelectric current obtained is directly proportional to the number of photo-electrons emitted. On increasing the frequency of incident light (or) by increasing the anode potential the speed of photo-electrons emitted changes but they have no effect on the number of electrons emitted. But by increasing the intensity of the incident light more photo-electrons will be emitted and thereby increasing the photoelectric current. 
    Also, if we increase the area on which the light is incident we can get more electrons emitted which increases the photoelectric current.
     

     

     

  • Question 8
    1 / -0.25

    Two electrons are moving with the same speed v. One electron enters a region of uniform electric field while the other enters a region of uniform magnetic field, then after sometime if the de-Broglie wavelengths of the two are l1  and l2  then :

  • Question 9
    1 / -0.25

    An electron in hydrogen atom first jumps from second excited state to first excited state and then, from first excited state to ground state. Let the ratio of wavelength, momentum and energy of photons in the two cases by x, y and z, then select the wrong answers :

    Solution

    • 1/λ=R[(1/nf2 )-(1/nl2 )
    • Therefore, for 3->2 , we get,
    • 1/λ1=R[(1/22 )-(1/32 )]=R[(1/4)-(1/9)]=5R/36
    • Therefore, λ1 =36/5R
    • Similarly, for 2->1, we get,
    • 1/λ2 =R[(1/12 )-(1/22 )]=R[(1-(1/4)]=3R/4
    • Therefore, λ2 =4/3R
    • Therefore, the ratio is,
    • x=λ1 /λ2 =(36/5R) x (3R/4)=27/5
    • hence, option B is wrong.
    •  
    • Now, the momentum is given as,
    • p=h/λ
    • Therefore, 3->2
    • p1 =h/λ1 =5Rh/36
    • similarly, for 2->1, we get,
    • p2 =h/λ2 =3Rh/4
    • therefore, the ratio is,
    • y=p1 /p2 =(5Rh/36)x(4/3Rh)=5/27
    • hence, option C is correct.
    • Now the energy difference between the two levels is,
    • ΔE=En-1 -En
    • Therefore, for 3->2
    • ΔE1 =E2 -E3 =(E1/22 )-(E1/32 )=E1 [(1/4)-(1/9)]=5E1 /36
    • Similarly, 2->1
    • ΔE2 =E1 -E2 =E1 -(E1/22 )=E1 [1-(1/4)]=3E1 /4
    • Therefore, the ratio is,
    • z= ΔE1 /ΔE2 =(5E1 /36) X (4/3E1 )=5/27
    • hence, option A and D are both correct.
    • therefore, we can say that the wrong is option (B)
       

  • Question 10
    1 / -0.25

    An electron is in an excited state in hydrogen-like atom. It has a total energy of –3.4 eV. If the kinetic energy of the electron is E and its de-Broglie wavelength is l, then

    Solution

    The potential energy of an electron is twice the kinetic energy(with negative sign) of an electron.
    PE=2E
    The total energy is: TE=PE+KE
                     −3.4=−2 ×3.4+KE
                     KE=3.4eV
    Let p be the momentum of an electron and m be the mass of an electron.
    E=p2 ​/2m
    p=√2 ​mE
     
    Now, the De-Broglie wavelength associated with an electron is:
    λ=h/p ​
    λ=h/√2 ​mE ​
    λ=6.6 ×1034 ​/√2 ​×9.1 ×10−31 ×(−3.4)×1.6 ×10−19
    λ=6.6 ×1034 ​ /9.95 ×10−25
    0 λ=0.66 ×10−9
    λ=6.6 ×10−10 m

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