Physics, asked by pv8465683, 11 months ago

The de Broglie wavelength associated with an electron accelerated
through a potential difference of 64 V

Answers

Answered by harsharora111

1

Answer:

Lemda = 12.27√V

Lemda = 12.27√8

⏩Lemda = 1.53 A° ⏪

Lemda = 1.53 × 10^-10m

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