Question

The mass of an electron is $9.1\times {10}^{-31}kg.IfitsK.E.is3.0\times {10}^{-25}J$. Calculate its wavelength

• A. 986.7 nm
• B. 896.7 nm
• C. 776.8 nm
• D. 1076.8 nm

Answer 1

The correct option is B

896.7 nm

De-Broglie wave lenght

$\lambda$ = $\frac{h}{p}$ $\Rightarrow$ $\lambda$ = $\frac{h}{mv}$

Now, Since $K.E=\frac{1}{2}m{v}^2$

$\therefore v={\left(\frac{2K.E}{m}\right)}^{\frac{1}{2}}={\left(\frac{2\times 3.0\times {10}^{-25}kg.m^2{s}^{-2}}{9.1\times {10}^{-31}kg}\right)}^{\frac{1}{2}}=812m{s}^{-1}$

$\lambda =\frac{h}{mv}=\frac{6.626\times {10}^{-34}Js}{(9.1\times {10}^{-31}kg)\left(812m{s}^{-1}\right)}$

$=8967\times {10}^{-10}m=896.7nm$