Calculate de Broglie wavelength of an electron travelling at `1 %` of the speed of light.

Options

A. `2.73 xx 10^-24`
B. `2.4 xx 10^-10`
C. `242.2 xx 10^10`
D. None of these

Answer

B

Explanation

(b) One percent of the speed of light is
`v = ((1)/(100)) (3.00 xx 10^9 ms^-1)`
=`(3.00 xx 10^8 ms^-1)`
Momentum of the electron `(p) = mv`
=`(9.11 xx 10^-31 kg)(3.00 xx 106 ms^-1)`
=`2.73 xx 10^-24 kg ms^-1`
The de-Broglie wavelength of this electron is
`lamda = (h)/(p) = (6.626 xx 10^-34)/(2.73 xx 10^-24 kgms^-1)`
`lamda = 2.424 xx 10^-10 m`.