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⚛️ Physics  ·  Dual Nature of Radiation and Matter  ·  NEET & JEE

An electron and a proton are accelerated through the same potential difference. What is the ratio of their de Broglie wavelengths, &lambda;<sub>e</sub>/&lambda;<sub>p</sub>?

Answer: 43.

  • A 1/&radic;1836
  • B 1/43
  • C 1836
  • D 43

Correct answer: D. 43

Explanation: For a particle of charge magnitude e accelerated through potential V, p = &radic;(2meV). Hence &lambda; = h/&radic;(2meV), so &lambda; is proportional to 1/&radic;m. Therefore &lambda;<sub>e</sub>/&lambda;<sub>p</sub> = &radic;(m<sub>p</sub>/m<sub>e</sub>) = &radic;1836 &asymp; 43.

evacuated tubelightmetal platee⁻e⁻e⁻collectorAbattery (variable)

In the photoelectric setup, light striking the metal plate ejects electrons that cross the evacuated tube to the collector, producing a measurable current on the ammeter.

Concept context

Photoelectric effect, de Broglie waves, Bohr's model, atomic spectra.

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