Question

If the de Broglie wavelength of an electron is equal to $10^{-3}$ times the wavelength of a photon of frequency $6\times 10^{14}Hz$ , then the speed of the electron is equal to (Speed of light $=3\times 10^{8}m{s}^{-1},$ Planck's constant $=6.63\times 10^{-34}Js,$ Mass of electron $=9.1\times 10^{-31}kg$ )

• A. $1.45\times 10^{6}m{s}^{-1}$
• B. $1.1\times 10^{6}m{s}^{-1}$
• C. $1.7\times 10^{6}m{s}^{-1}$
• D. $1.8\times 10^{6}m{s}^{-1}$

Answer 1

Answer: (A)

${\lambda }_e=10^{-3}{\lambda }_p$
$f_p=6\times 10^{14}Hz=\frac{C}{{\lambda }_p}$
${\lambda }_p=\frac{3\times 10^8}{6\times 10^{14}}=0.5\times 10^{-6}m$
${\lambda }_e=0.5\times 10^{-9}=\frac{h}{m_e{v}_e}$
$v_e=\frac{6.63\times 10^{-34}}{9.1\times 10^{-31}\times 0.5\times 10^{-9}}$
$=1.45\times 10^{6}m{s}^{-1}$