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Test: Dual Nature of Radiation of Matter- 2 (March 8) - NEET MCQ


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10 Questions MCQ Test - Test: Dual Nature of Radiation of Matter- 2 (March 8)

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Test: Dual Nature of Radiation of Matter- 2 (March 8) - Question 1

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 for Test: Dual Nature of Radiation of Matter- 2 (March 8) - Question 1

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

Test: Dual Nature of Radiation of Matter- 2 (March 8) - Question 2

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 for Test: Dual Nature of Radiation of Matter- 2 (March 8) - Question 2

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%

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Test: Dual Nature of Radiation of Matter- 2 (March 8) - Question 3

When a photon of light collides with a metal surface, number of electrons, (if any) coming out is

Detailed Solution for Test: Dual Nature of Radiation of Matter- 2 (March 8) - Question 3

When photon strikes with the electron it completely transfers it’s energy to the electron as during photoelectric experiment the threshold frequency required is used by the electron to eject from the atom which is also called the work function and the remaining energy in electron is kinetic energy which can be measured. Now (before collision) since photon is a particle it must have mass and thus it has energy equivalent to E=mc2 .
After collision when it completely transfers it’s energy to electron thus E=0
And therefore 0=mc2 thus i guess photon vanishes.
 

Test: Dual Nature of Radiation of Matter- 2 (March 8) - Question 4

A point source of light is used in photoelectric effect. If the source is removed farther from the emitting metal, the stopping potential

Detailed Solution for Test: Dual Nature of Radiation of Matter- 2 (March 8) - Question 4

When the source is moved away from the emitter, intensity of the incident radiation decreases but frequency remains the same so there will be no change in the stopping potential. Thus, it remains constant.

Test: Dual Nature of Radiation of Matter- 2 (March 8) - Question 5

The maximum kinetic energy of photoelectrons emitted from a surface when photons of energy 6 eV fall on it is 4eV. The stopping potential is Volts is

Detailed Solution for Test: Dual Nature of Radiation of Matter- 2 (March 8) - Question 5

Given, the maximum kinetic energy: Kmax​=4eV
If V0​ be the stopping potential, then Kmax​=eV0​
⇒eV0​=4eV 
⇒V0​=4V

Test: Dual Nature of Radiation of Matter- 2 (March 8) - Question 6

Cut off potentials for a metal in photoelectric effect for light of wavelength l1, l2 and l3 is found to be V1, V2 and V3 volts if V1, V2 and V3 are in Arithmetic Progression and l1, l2 and l3 will be

Detailed Solution for Test: Dual Nature of Radiation of Matter- 2 (March 8) - Question 6

We know that,
eV=(hc/λ)-w
V=(hc/eλ)-(w/e)
Arithmetic progression =>V2=(V1+V2)/2
Now,
(hc/eλ2)-w/e=1/2[(hc/eλ1)-(w/e) +(hc/eλ3) -(w/e)]
=>1/ λ2=1/2[(1/ λ1)+(1/λ3)]
=>2/ λ2=1/ λ1 + 1/λ3
Hence the correct answer is harmonic Progression.

Test: Dual Nature of Radiation of Matter- 2 (March 8) - Question 7

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 for Test: Dual Nature of Radiation of Matter- 2 (March 8) - Question 7

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.

Test: Dual Nature of Radiation of Matter- 2 (March 8) - Question 8

In a photoelectric experiment, electrons are ejected from metals X and Y by light of intensity I and frequency f. The potential difference V required to stop the electrons is measured for various frequencies. If Y has a greater work function than X ; which one of the following graphs best illustrates the expected results ?

Detailed Solution for Test: Dual Nature of Radiation of Matter- 2 (March 8) - Question 8

The answer is option A. 
First, the gradient of the graph cannot change (always = h/e ), so answers are A or D.
If Y has a greater work function than X, then the graph for Y should have a more negative y-intercept. therefore figure in option A depicts the concept.

Test: Dual Nature of Radiation of Matter- 2 (March 8) - Question 9

A image of the sun of formed by a lens of focal-length of 30 cm on the metal surface of a photo-electric cell and a photo-electrci current is produced. The lens forming the image is then replaced by another of the same diameter but of focal length 15 cm. The photo-electric current in this case is

Detailed Solution for Test: Dual Nature of Radiation of Matter- 2 (March 8) - Question 9

Lenses of the same diameter collect equal amounts of light.
Intensity is the measure of amount of light collected, hence the intensity remains the same.
Intensity is measured by the photoelectric current. So the photoelectric current would also remain the same.

Test: Dual Nature of Radiation of Matter- 2 (March 8) - Question 10

An electron with initial kinetic energy of 100eV is acceleration through a potential difference of 50V. Now the de-Broglie wavelength of electron becomes.

Detailed Solution for Test: Dual Nature of Radiation of Matter- 2 (March 8) - Question 10

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