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 $=6.63\times 10^{- 34}J \, s,$ Mass of electron $=9.1\times 10^{- 31} \, kg$ )

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

Answer 1

Answer: (A)

$\lambda _{e}=10^{- 3}\lambda _{p}$
$f_{p}=6\times 10^{14}Hz=\frac{C}{\lambda _{p}}$
$\lambda _{p}=\frac{3 \times 10^{8}}{6 \times 10^{14}}=0.5\times 10^{- 6}m$
$\lambda _{e}=0.5\times 10^{- 9}=\frac{h}{m_{e} v_{e}}$
$v_{e}=\frac{6.63 \times 10^{- 34}}{9.1 \times 10^{- 31} \times 0.5 \times 10^{- 9}}$
$=1.45\times 10^{6}ms^{- 1}$