Question

The de-broglie wavelength of an electron that has been accelerated through a potential difference of $100V$ is:

• A. $0.1226$ nm
• B. $1.226$ nm
• C. $12.26$ nm
• D. $0.01226$ nm

Answer 1

The correct option is A $0.1226$ nm

We know that, $\frac{1}{2}m{v}^2=K.E.=$ev

Charge on electron: $1.6\times {10}^{-19}$ coulombs

Mass of electron$=9.1\times {10}^{-31}$ kg

$\frac{1}{2}\times 9.1\times {10}^{-31}\times v^2=1.6\times {10}^{-19}\times 100$

$v^2=2\times 1.6\times {10}^{-9}\times \frac{100}{9.1\times {10}^{-31}}$

$v^2=\sqrt{0.0150}$

$v=1.88\times {10}^6$m/s

Now, $\lambda =\frac{h}{mv}$

$\therefore \lambda =0.1226$nm.