Question

The mass of an eletron is $9.1\times {10}^{-31}$ kg. If its kinetic energy is $3.0\times {10}^{-25}$J. Calculate its de-broglie wavelength.

• A. $812\text{nm}$
• B. $897\text{nm}$
• C. $880\text{nm}$
• D. $910\text{nm}$

Answer 1

The correct option is B $897\text{nm}$

We know Kinetic energy (K.E) =$\frac{1}{2}m{v}^2$
$=\frac{1}{2}m{v}^2=K.E.$
$\Rightarrow v={\left(\frac{2K.E}{m}\right)}^{\frac{1}{2}}$
Putting the values according to the given data,
$v={\left(\frac{2\times 3\times {10}^{-25}kg{m}^2{s}^{-2}}{9.1\times {10}^{-31}kg}\right)}^{\frac{1}{2}}$
$\Rightarrow v=812m{s}^{-1}$
According to de-broglie wavelength
$\lambda =\frac{h}{mv}$
$=\frac{6.626\times {10}^{-34}}{(9.1\times {10}^{-31}kg)\left(812m{s}^{-1}\right)}$
$=8967\times {10}^{-10}m=896.7nm$