Question

Calculate the de Broglie wavelength of an electron having kinetic energy of $1.6\times {10}^{-6}erg$ $(m_e=9.11\times {10}^{-28}g,h=6.62\times {10}^{-27}erg-sec)$.

• A. $90A^o$
• B. $0.00029A^o$
• C. $0.0122A^o$
• D. $1.29A^o$

Answer 1

Answer: (C)

$0.0122A^o$

The expression for the kinetic energy E is $E=\frac{1}{2}m{u}^2$
Hence, $u=\sqrt{2E}m$......(1)
The De-broglie equation is $\lambda =\frac{h}{mu}$.....(2)
Substittue (2) in (1)
$\lambda =\frac{h}{\sqrt{2mE}}$
Substitute values in the above expression.
$\lambda =\frac{6.62\times {10}^{-27}}{\sqrt{2\times 9.11\times {10}^{-28}\times 1.6\times {10}^{-6}}}=1.226\times {10}^{-10}cm=0.01226A^o$
Hence, the de Broglie wavelength of the electron is $0.01226A^o$