Verified Solution • 2025–26 Session moderate

QJEEM16Chf89d: JEE Main 2016 Chemistry - Class 11 JEEM16Chf89d Solution: A stream of electrons from a heated filament was passed between …

A stream of electrons from a heated filament was passed between two charged plates kept at a potential difference V esu. If e and m are charge and mass of an electron, respectively, then the value of h / λ\lambda (where λ\lambda is wavelength associated with electron wave) is given by :

Multiple Choice Options

  • A 2meV
  • B meV\sqrt {meV}
  • C 2meV\sqrt {2meV}
  • D meV

Step-by-Step Verified Solution

We know, Kinetic Energy (KE) = 12mv2{1 \over 2}m{v^2}

∴\therefore mv2 = 2KE

⇒\Rightarrow m2v2 = 2mKE

⇒\Rightarrow mv = 2mKE\sqrt {2m{K_E}}

For an electron of charge 'e' which is passes through 'V' volt, kinetic energy of electron will be

KE = eV

∴\therefore m = 2meV\sqrt {2meV}

We know de-Broglie wavelength for an electron (λ)\left( \lambda \right) = hp=hmv{h \over p} = {h \over {mv}}

∴\therefore hλ{h \over \lambda } = mv = 2meV\sqrt {2meV}

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