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de-Broglie wavelength associated with an electron accelerated through a potential difference V is `lamda`. What will be its wavelength when the accelerating potential is increased to 4 V?

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As `lamdaprop(1)/(sqrtV)`, hence on increasing accelerating potential from V to 4V, de-Broglie wavelength decreases from `lamda` to `(lamda)/(2)`.
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