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| 1. |
de-Broglie wavelength of an electron wave is 1.227 Å, when the electron is accelerated by a potential difference of _____. |
| Answer» Solution :`100V.` [As `LAMDA=(1.227)/(sqrtV)nm=(12.27)/(SQRT(V))Å`, hence `lamda` is 1.2227 Å only when V=100 VOLTS]. | |