Question

Electrons are accelerated through a p.d. of 150V. Given $m=9.1\times {10}^{-31}$ Js, the de Broglie wavelength associated with it is

• A. 1.5$A^0$
• B. 1.0$A^0$
• C. 3.0 $A^0$
• D. 0.5 $A^0$

Answer 1

Answer: (B)

De-Brogtie wavelength
$\lambda =\frac{h}{mv}$

And
$\frac{1}{2}m{v}^2=eV$

$mv=\sqrt{2meV}$

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

$\lambda =\frac{h}{\sqrt{2meV}}$

$\lambda =\frac{6.62\times {10}^{-34}}{\sqrt{2\times 9.1\times {10}^{-31}\times 1.6\times {10}^{-19}\times 150}}$

$\lambda =1.0A^0$

So, the answer is option (B).