Doubts
Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents
Sign in to save progress →
Doubts
Sign in

Dual Nature of Radiation question

2019 · 8 Apr · Shift 1 · Q48
Guest · try one question or challengeGuest · try one question or challenge
  1. PYQ Library
  2. /JEE Main
  3. /Physics
  4. /Dual Nature of Radiation
  5. /2019 · 8 Apr · Shift 1 · Q48

Dual Nature of Radiation question

2019 · 8 Apr · Shift 1 · Q48

JEE MainPhysicsDual Nature of RadiationMCQ+4 / −1
Two particles move at right angle to each other. Their de-Broglie wavelengths are λ1\lambda _1λ1​ and λ2\lambda _2λ2​ respectively. The particles suffer perfectly inelastic collision. The de-Broglie wavelength λ2\lambda _2λ2​ of the final particle, is given by :
  1. A
    λ=λ1+λ22\lambda = {{{\lambda _1} + {\lambda _2}} \over 2}λ=2λ1​+λ2​​
  2. B
    1λ2=1λ12+1λ22{1 \over {{\lambda ^2}}} = {1 \over {\lambda _1^2}} + {1 \over {\lambda _2^2}}λ21​=λ12​1​+λ22​1​
  3. C
    λ=λ1λ2\lambda = \sqrt {{\lambda _1}{\lambda _2}}λ=λ1​λ2​​
  4. D
    2λ=1λ1+1λ2{2 \over \lambda } = {1 \over {{\lambda _1}}} + {1 \over {{\lambda _2}}}λ2​=λ1​1​+λ2​1​
PreviousNext

More from Dual Nature of Radiation

  • A nucleus A, with a finite de-broglie wavelength λ A, undergoes spontaneous fission into two nuclei B and C of equal mass. B flies in the same direction as that of A, while C flies in the opposite direction with a velocity equal to…2019 · MCQ
  • The electric field of light wave is given as E=10−3cos(5×10−72πx​−2π×6×1014t)x∧CN​ This light falls on…2019 · MCQ
  • A particle 'P' is formed due to a completely inelastic collision of particles 'x' and 'y' having de-Broglie wavelengths 'λ x' and 'λ y' respectively. If x and y were moving in opposite directions, then the de-Broglie…2019 · MCQ
  • Surface of certain metal is first illuminated with light of wavelength λ 1 = 350 nm and then, by light of wavelength λ 2 = 540 nm. It is found that the maximum speed of the photo electrons in the two cases differ by a…2019 · MCQ
  • The magnetic field associated with a light wave is given, at the origin, by B = B0 [sin(3.14 × 107)ct + sin(6.28 × 107)ct]. If this light falls on a silver plate having a work function of 4.7 eV, what will be the maximum…2019 · MCQ
  • In a photoelectric effect experiment the threshold wavelength of the light is 380 nm. If the wavelentgh of incident light is 260 nm, the maximum kinetic energy of emitted electrons will be: Given E (in eV) = 1237/λ (in nm)2019 · MCQ
  • A 2 mW laser operates at wavelength of 500 nm. The number of photons that will be emitted per second is : [Given Planck's constant h = 6.6 × 10–34 Js, speed of light c = 3.0 × 108 m/s]2019 · MCQ
  • In an electron microscope, the resolution that can be achieved is of the order of the wavelength of electrons used. To resolve a width of 7.5 × 10–12 m, the minimum electron energy required is close to -2019 · MCQ