Question

An electron has kinetic energy $2.8 \times 10^{-23}$ J. de-Broglie wavelength will be nearly $(m_{e}= 9.1 \times 10^{-31}$ kg)

• A. $9.28 \times 10^{-4}\, m$
• B. $9.28\times 10^{-7}\, m$
• C. $9.28 \times10^{-8}\, m$
• D. $9.28 \times 10^{-10}\, m$

Answer 1

Answer: (C)

Formula for de-Broglie wavelength is
$\lambda=\frac{h}{p} or \lambda =\frac{h}{mv} \Rightarrow eV =\frac{1}{2} mv^{2} or v=\sqrt{\frac{2eV}{m}}$
$\lambda=\frac{h}{\sqrt{2me V}}=\frac{6.62\times10^{-34}}{\sqrt{2\times9.1\times10^{-31}\times2.8\times10^{-23}}}$
$\lambda =9.28 \times10^{-8} meter.$