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MCQ Practice Test & Solutions: Chapter Test: Dual Nature of Matter and Radiation - 2 (30 Questions)

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Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 60 minutes
  • - Number of Questions: 30

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Chapter Test: Dual Nature of Matter and Radiation - 2 - Question 1


Detailed Solution: Question 1


Chapter Test: Dual Nature of Matter and Radiation - 2 - Question 2


Detailed Solution: Question 2


Chapter Test: Dual Nature of Matter and Radiation - 2 - Question 3

When a metal surface is illuminated by light of wavelengths 400 nm and 250 nm, the maximum velocities of the photoelectrons ejected are v and

2v respectively. The work function of the metal is - (h = Planck's constant, c = velocity of light in air)

Detailed Solution: Question 3

Chapter Test: Dual Nature of Matter and Radiation - 2 - Question 4

When light of wavelength 300 nm (nanometer) falls on a photoelectric emitter, photoelectrons are liberated. For another emitter, however light of 600nm wavelength is sufficient for creating photoemission. What is the ratio of the work functions of the two emitters ?

Detailed Solution: Question 4


Chapter Test: Dual Nature of Matter and Radiation - 2 - Question 5


Detailed Solution: Question 5


Chapter Test: Dual Nature of Matter and Radiation - 2 - Question 6

The threshold frequency for a certain metal is v0. When light of frequency V = 2v0 is incident on it, the maximum velocity of photoelectrons is 4 × 106 m/s. If the frequency of incident radiations is increased to 5v0, then the maximum velocity of photoelectrons in m/s will be -

Detailed Solution: Question 6


Chapter Test: Dual Nature of Matter and Radiation - 2 - Question 7


Detailed Solution: Question 7


Chapter Test: Dual Nature of Matter and Radiation - 2 - Question 8

When radiation of wavelength λ is incident on a metallic surface, the stopping potential is 4.8 volts. If the same surface is illuminated with

radiation of double the wavelength, then the stopping potential becomes 1.6 volts. Then the threshold wavelength for the surface is -

Detailed Solution: Question 8


Chapter Test: Dual Nature of Matter and Radiation - 2 - Question 9

If K1 and K2 are the maximum kinetic energies of photoelectrons emitted when lights of wavelength λ1 and λ2 respectively incident on a metallic surface and λ1 = 3λ2 , then -

Detailed Solution: Question 9


Chapter Test: Dual Nature of Matter and Radiation - 2 - Question 10


Detailed Solution: Question 10


Chapter Test: Dual Nature of Matter and Radiation - 2 - Question 11


Detailed Solution: Question 11


Chapter Test: Dual Nature of Matter and Radiation - 2 - Question 12


Detailed Solution: Question 12


Chapter Test: Dual Nature of Matter and Radiation - 2 - Question 13

The figure shows the variation of photo current with anode potential for a photo-sensitive surface for three different radiations. Let Ia, Ib and Ic be the intensities and fa, fb and fc be the frequencies for the curves a, b and c respectively -


Detailed Solution: Question 13

Chapter Test: Dual Nature of Matter and Radiation - 2 - Question 14


Detailed Solution: Question 14


Chapter Test: Dual Nature of Matter and Radiation - 2 - Question 15

In the following diagram if V2 > V1 then -


Detailed Solution: Question 15


Chapter Test: Dual Nature of Matter and Radiation - 2 - Question 16

If the distance of a 100 watt lamp is increased from a photo cell, the saturation current I in the photocell varies with the distance d as -

Detailed Solution: Question 16


Chapter Test: Dual Nature of Matter and Radiation - 2 - Question 17


Detailed Solution: Question 17

Chapter Test: Dual Nature of Matter and Radiation - 2 - Question 18


Detailed Solution: Question 18

Chapter Test: Dual Nature of Matter and Radiation - 2 - Question 19


Detailed Solution: Question 19


Chapter Test: Dual Nature of Matter and Radiation - 2 - Question 20


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Chapter Test: Dual Nature of Matter and Radiation - 2 - Question 21


Detailed Solution: Question 21


Chapter Test: Dual Nature of Matter and Radiation - 2 - Question 22

The stopping potential (V0) versus frequency (v) plot of a substance is shown in figure, the threshold wavelength is –


Detailed Solution: Question 22


Chapter Test: Dual Nature of Matter and Radiation - 2 - Question 23

Maximum velocity of the photoelectrons emitted by a metal surface is 1.2 × 106 ms–1 . Assuming the specific charge of the electron to be 1.8 × 1011

C Kg–1, the value of the stopping potential in volt will be -

Detailed Solution: Question 23


Chapter Test: Dual Nature of Matter and Radiation - 2 - Question 24


Detailed Solution: Question 24


Chapter Test: Dual Nature of Matter and Radiation - 2 - Question 25

A cesium photocell, with a steady potential difference of 60 V across, is illuminated by a bright point source of light 50 cm away. When the same light is placed 1 m away, the photoelectrons emitted from the cell -

Detailed Solution: Question 25


Chapter Test: Dual Nature of Matter and Radiation - 2 - Question 26

Let nr and nb be respectively the number of photons emitted by a red bulb and a blue bulb of equal power in a given time.

Detailed Solution: Question 26

Since, Pr​=Pb​ 
r for red and b for blue.
Pr​=Pb​
or, nr​× (hc/λr)  ​=nb​× (hc​/λb)​
or,  ​(nr/nb)​​= λr​​/λb
Since, the wavelength of red bulb is greater than the wavelength of blue bulb.
or,  nr​>nb

Chapter Test: Dual Nature of Matter and Radiation - 2 - Question 27

10-3 W of 5000 Å light is directed on a photoelectric cell. If the current in the cell is 0.16 mA, the percentage of incident photons which produce photoelectrons, is

Detailed Solution: Question 27

Current is 0.16×10−6 Amp it means 0.16×10−6 Coulomb charge is flowing per second
So, n=0.16×10−6C /1.6×10−19C ​=1012 electrons are generated per second
Now we notice that one photon has energy E, E=hc/λ=​=(6.62×10−34Js×3×108ms-1)/(5000×10−10m) ​=3.972×10−19Joule
So, number of photon in 10−3W will be N=10−3/3.972×10−19 ​=0.25×1016 this is number of photons incident per second
So required percentage is (n/N)×100=1014/(0.25×1016) ​=0.04%

Chapter Test: Dual Nature of Matter and Radiation - 2 - Question 28

The stopping potential for the photo electrons emitted from a metal surface of work function 1.7eV is 10.4 V. Identify the energy levels corresponding to the transitions in hydrogen atom which will result in emission of wavelength equal to that of incident radiation for the above photoelectric effect

Detailed Solution: Question 28

As we know that the stopping potential of the photoelectron is equal to the maximum kinetic energy of the photoelectron,
KEmax​=10.4V
Now, in photoelectric effect,
Energy of incident radiation (Ein​) = work function + K.Emax​
⇒ Ein​=1.7+10.4
⇒ Ein​=12.1eV
Now, for 0 hydrogen atom,
Energy of first energy level, E1​=−13.6eV
Energy of second energy level, E2​=−3.4eV
Energy of third energy level, E3​=−1.5eV
Hence, a transition from third to first energy level will result in emission of radiation of energy = E3​−E1​=12.1eV which is same as the energy of incident radiation of above photoelectric effect.
Thus, correct answer is n=3 to 1
 

Chapter Test: Dual Nature of Matter and Radiation - 2 - Question 29

A point source causes photoelectric effect from a small metal plate. Which of the following curves may represent the saturation photocurrent as a function of the distance between the source and the metal?

Detailed Solution: Question 29

Saturation current is the maximum current possible and it will be directly proportional to the number of number of electrons falling on collector plate per second which depend on number of photons
 incident on the cathode as one photon contribute in one electron and the number of photons is actually
 proportional to intensity which varies 
As intensity I∝1/r2​, where r is the distance 
So the correct graph will be decreasing with power2 of distance and it will be rapidly decreasing with a higher value of r.
 

Chapter Test: Dual Nature of Matter and Radiation - 2 - Question 30

Photons with energy 5eV are incident on a cathode C, on a photoelectric cell. The maximum energy of the emitted photoelectrons is 2eV. When photons of energy 6eV are incident on C, no photoelectrons will reach the anode A if the stopping potential of A relative to C is

Detailed Solution: Question 30

When 5eV is incident the kinetic energy is 2eV it simply means the work function is W=5eV−2eV=3eV
Similarly, when 6eV is incident the kinetic energy should be   6eV−W=6eV−3eV=3eV
 it simply means to stop them we need a negative potential at anode equal to 3eV/e​=3V
So, the answer is −3V i.e. option B is correct.

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