Question

What is de-Broglie wavelength of electron having energy 10 keV?

• A. $0.12\stackrel{\text{o}}{A}$
• B. $1.2\stackrel{\text{o}}{A}$
• C. $12.2\stackrel{\text{o}}{A}$
• D. None of these

Answer 1

Answer: (A)

de-Broglie wavelength of a particle is given by $\lambda =\frac{h}{mv}$ ... (i)
where $h$ is Plancks constant. If kinetic energy of particle of mass m is $v$ , then $K=\frac{1}{2}m{v}^2$
$\Rightarrow$ $v=\sqrt{\frac{2K}{m}}$ ... (ii)
Combining Eqs. (i) and (ii), we get
$\lambda =\frac{h}{m\sqrt{\frac{2K}{m}}}=\frac{h}{\sqrt{2mK}}$ ... (iii)
Given: $m=9.1\times {10}^{-31}kg$
$K=10keV=10\times {10}^3\times 1.6\times {10}^{-19}J$
$h=6.6\times {10}^{-34}J-s$
Substituting the above values in Eq. (iii), we get
$\lambda =\frac{6.6\times {10}^{-34}}{\sqrt{2\times 9.1\times {10}^{-31}\times 10\times {10}^3\times 1.6\times {10}^{-19}}}$
$=1.22\times {10}^{-11}$
$\approx 0.12\stackrel{\text{o}}{A}$
Note: If an electron is accelerated through a potential difference of $V$ volt, then Eq. (iii) takes the form $\lambda =\frac{h}{\sqrt{2meV}}$ After putting the numerical values for electrons, we get $\lambda =\sqrt{\frac{150}{V}}\stackrel{\text{o}}{A}$