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Test: Dual Nature of Radiation and Matter - 2 - CUET Humanities MCQ


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10 Questions MCQ Test Agriculture Practice Tests: CUET Preparation - Test: Dual Nature of Radiation and Matter - 2

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

Work function of a metal is 5.2 × 10–18. Its threshold wavelength will be

Detailed Solution for Test: Dual Nature of Radiation and Matter - 2 - Question 1

Work function W = hc/λ
λ = hc/W

 = 376.4Å

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

Which one of the following graphs represents the variation of maximum kinetic energy (Ek) of the emitted electrons with frequency (v) in photoelectric effect correctly?

Detailed Solution for Test: Dual Nature of Radiation and Matter - 2 - Question 2

According to photoelectric equation,
Ek = hv — hv0
From the equation, we can conclude that the graph of Ek vs v is a straight line with positive slope (h) and negative intercept (hv0).
∴ Graph (4) represents the variation of Ek with v correctly.

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

The work functions of metals A, B and C are 1.92 eV, 2.0 eV and 5 eV, respectively. According to Einstein's equation, which metal(s) will emit the photoelectrons for a radiation of wavelength 4100 Å?

Detailed Solution for Test: Dual Nature of Radiation and Matter - 2 - Question 3

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

There are n1 photons of frequency v1 in a beam of light. In an equally energetic beam, there are n2 photons of frequency v2. The correct relation is

Detailed Solution for Test: Dual Nature of Radiation and Matter - 2 - Question 4

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

A metal has work function 3.31 eV. It is illuminated by light of wavelength 3 x 10-7m.What is the threshold frequency for photoelectric emission? ( Take h = 6.62 ×10-34 J s)

Detailed Solution for Test: Dual Nature of Radiation and Matter - 2 - Question 5

Work function is given by
W = hv0

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

Threshold wavelength for photoelectric emission, from a metal surface, is 5200 Å. Photoelectrons will be emitted when this surface is illuminated with monochromatic radiation from

Detailed Solution for Test: Dual Nature of Radiation and Matter - 2 - Question 6

λ0 = 5200 × 10− 10 m
Photo electrons will be emitted if the incident radiations has a wavelength less than λ0
∴ UV lamp emits radiation of wavelength less than 5200 × 10− 10 m

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

A radio transmitter operates on a wavelength of 1500 m at a power of 400 kilowatt. The energy of the radio photon (in joules) is

Detailed Solution for Test: Dual Nature of Radiation and Matter - 2 - Question 7

According to Planck's quantum theory, a source of radiation emits energy in the form of photons, which travel in straight line.
The energy of photon is E = hv = hc/λ 
Hence, energy of radio photon is E = 6.6 × 10-34 × 2 × 105 J

∴ E = 1.32 × 10-28 J

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

A metal whose work function is 3.31 eV is illuminated by light of wavelength 3  10-7 m. What is the threshold frequency for photoelectric emission? (Take h = 6.62 x 10-34 Js)

Detailed Solution for Test: Dual Nature of Radiation and Matter - 2 - Question 8

The threshold frequency v0 is given by 
v0 = W0/h

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

A light having wavelength 300 nm falls on a metal surface. Work function of metal is 2.54 eV. What is the stopping potential?

Detailed Solution for Test: Dual Nature of Radiation and Matter - 2 - Question 9

Energy of photon = 1243/300 = 4.14eV 
work function = 2.54eV
Kinetic energy = 4.14-2.54 = 1.60eV
stopping potential = 1.6v

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

Relation between wavelength of photon and electron of same energy is

Detailed Solution for Test: Dual Nature of Radiation and Matter - 2 - Question 10

Wavelength  λ = hc/p
 
Mass of photon is less than that of electron.
Wavelength is inversely proportional to the mass.
 Wavelength of photon is greater than that of electron.
Hence, λph > λe.

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