Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Dual Nature of Radiation question

2023 · 30 Jan · Shift 2 · Q49
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Physics
  4. /Dual Nature of Radiation
  5. /2023 · 30 Jan · Shift 2 · Q49

Dual Nature of Radiation question

2023 · 30 Jan · Shift 2 · Q49

JEE MainPhysicsDual Nature of RadiationMCQ+4 / −1
An electron accelerated through a potential difference V1V_{1}V1​ has a de-Broglie wavelength of λ\lambdaλ. When the potential is changed to V2V_{2}V2​, its de-Broglie wavelength increases by 50%50 \%50%. The value of (V1V2)\left(\frac{V_{1}}{V_{2}}\right)(V2​V1​​) is equal to
  1. A
    32\frac{3}{2}23​
  2. B
    4
  3. C
    3
  4. D
    94\frac{9}{4}49​
View written solutionFree

Correct answer: D

  1. For an electron accelerated through potential difference VVV, its kinetic energy is eV=p22meV = \frac{p^2}{2m}eV=2mp2​ So momentum p=2meVp = \sqrt{2meV}p=2meV​

  2. The de-Broglie wavelength is λ=hp=h2meV\lambda = \frac{h}{p} = \frac{h}{\sqrt{2meV}}λ=ph​=2meV​h​ Thus, λ∝1V\lambda \propto \frac{1}{\sqrt{V}}λ∝V​1​

  3. Initially, at potential V1V_1V1​, wavelength is λ\lambdaλ. When potential is changed to V2V_2V2​, wavelength increases by 50%50\%50%: λ2=1.5λ1=32λ1\lambda_2 = 1.5\lambda_1 = \frac{3}{2}\lambda_1λ2​=1.5λ1​=23​λ1​

  4. Using λ∝1V\lambda \propto \frac{1}{\sqrt{V}}λ∝V​1​, λ2λ1=V1V2\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{V_1}{V_2}}λ1​λ2​​=V2​V1​​​ Substitute λ2λ1=32\frac{\lambda_2}{\lambda_1} = \frac{3}{2}λ1​λ2​​=23​: 32=V1V2\frac{3}{2} = \sqrt{\frac{V_1}{V_2}}23​=V2​V1​​​

  5. Squaring both sides, V1V2=(32)2=94\frac{V_1}{V_2} = \left(\frac{3}{2}\right)^2 = \frac{9}{4}V2​V1​​=(23​)2=49​

  6. Checking options:

  • A: 32\frac{3}{2}23​ ❌
  • B: 444 ❌
  • C: 333 ❌
  • D: 94\frac{9}{4}49​ ✅

Therefore, the correct answer is D.

PreviousNext

More from Dual Nature of Radiation

  • If a source of electromagnetic radiation having power 15 kW produces 1016 photons per second, the radiation belongs to a part of spectrum is. (Take Planck constant h=6×10−34Js )2023 · MCQ
  • If the two metals A and B are exposed to radiation of wavelength 350 nm. The work functions of metals A and B are 4.8eV and 2.2eV. Then choose the correct…2023 · MCQ
  • When light of frequency twice the threshold frequency is incident on the metal plate, the maximum velocity of emitted electron is v1. When the frequency of incident radiation is increased to five times the threshold value, the maximum…2022 · Numerical
  • The light of two different frequencies whose photons have energies 3.8 eV and 1.4 eV respectively, illuminate a metallic surface whose work function is 0.6 eV successively. The ratio of maximum speeds of emitted electrons for the two…2022 · MCQ
  • A metal exposed to light of wavelength 800 nm and and emits photoelectrons with a certain kinetic energy. The maximum kinetic energy of photo-electron doubles when light of wavelength 500 nm is used. The…2022 · MCQ
  • The ratio of wavelengths of proton and deuteron accelerated by potential Vp and Vd is 1 : 2​. Then the ratio of Vp to Vd will be :2022 · MCQ
  • A proton, a neutron, an electron and an α particle have same energy. If λ p, λ n, λ e and λ a are the de Broglie's wavelengths of proton, neutron, electron and α particle respectively, then…2022 · MCQ
  • A parallel beam of light of wavelength 900 nm and intensity 100Wm−2 is incident on a surface perpendicular to the beam. The number of photons crossing 1 cm2 area perpendicular to the beam in one…2022 · MCQ