1.

Electrons are accelerated through a potential difference of 150V. Calculate thede Broglie wavelength.

Answer»

`lambda_3 = lambda_p`
`lambda_e lt lambda_p`
`lambda_e gt lambda_p`
`n_3 to n_1`

Solution :`(LAMBDA E)/(lambda p)=SQRT((MP)/(me)) = sqrt(1837)`


Discussion

No Comment Found