    Question

# The stream of electrons from a heated filament was passed between two charged plates kept at a potential difference of 2 V volts. If e and m are the charge and the mass of electron respectively, then the value of hλ is: (where λ is the de Broglie wavelength of the electron)

A
meV
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B
2meV
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C
2meV
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D
2meV
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Solution

## The correct option is B 2√meVGiven: Charge of electron = e Potential difference applied = 2V Mass of electron = m Let v be the veloctiy of the electron. Kinetic energy would be =12mv2 Kinetic energy gained by electron due to the applied potential =e(2V) ∴2eV=12mv2⟹2meV=12m2v2⟹4meV=(mv)2⟹2√meV=mv de Broglie wavelength of the electron given by: λ=hmv⟹hλ=mv Since mv=2√meV hλ=2√meV  Suggest Corrections  0      Similar questions  Explore more