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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 2meV
Given:
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=12mv22meV=12m2v24meV=(mv)22meV=mv

de Broglie wavelength of the electron given by:
λ=hmvhλ=mv
Since
mv=2meV
hλ=2meV

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