Question

The mass of an electron is $9.1\times {10}^{-31}kg$. If the K.E. is given as $3.0\times {10}^{-25}J$. Calculate the wavelength of the electron.

• A. $8967\stackrel{o}{A}$
• B. $6242\stackrel{o}{A}$
• C. $7264\stackrel{o}{A}$
• D. $6966\stackrel{o}{A}$

Answer 1

The correct option is A $8967\stackrel{o}{A}$

$K.E.=\frac{1}{2}m{v}^2$-----(1)
$\lambda =\frac{h}{mv}$
$v=\frac{h}{m\lambda }$ ----(2)
By eq. (1) and (2),
$K.E.=\frac{1}{2}m(\frac{h}{m\lambda }{)}^2$
$\lambda =\sqrt{\frac{h^2}{2mK.E}}$
$=\sqrt{\frac{(6.626\times {10}^{-34}{)}^2}{2\times 9.1\times {10}^{-31}kg\times 3.0\times {10}^{-25}J}}$
$=8967\times {10}^{-7}m=8967\stackrel{o}{A}$