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An electron is accelerated through a potential difference of `10,000V`. Its de Broglie wavelength is, (nearly): `(me=9xx10^(-31)kg)`A. 12.2nmB. `12.2xx10^(-13)m`C. `12.2xx10^(-12)m`D. `12.2xx10^(-14)m` |
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Answer» Correct Answer - C be Broglie wave length of electron `(lambda_(e))=(12.27)sqrt(v)A^(@)` `v=` accelerating voltage `lambda_(e)=(12.27)/sqrt(10000)xx10^(-10)m` `lambda_(e)=12.2xx10^(-12)m` |
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