The de Brogle wavelength of an electron accelrated to a potential of 400 V is approximately

Options

A. `0.03nm`
B. `0.04nm`
C. `0.12nm`
D. `0.06nm`

Answer

d

Explanation

de Brogliewavelength `lamda = -(h)/(p)= (h)/(sqrt2meV)`
Here , V= 400V
`lamda= (6.6xx(1^(-34))/(sqrt2xx9.1xx10^9-31)xx1.6xx10^(-19)xx400)`
`= (6.6xx10^(-10))/(sqrt2xx9.1xx1.6xx4)= 0.61xx10^(-10)m`
`0.06xx10^(-9)m=0.06nm`