Question

The de Broglie wavelength associated with an electron of energy 500 eV is given by

(take $h=6.63\times {10}^{-34}Js,m=9.11\times {10}^{-31}kg$ )

• A. 0.28$A^0$
• B. 1.410 $A^0$
• C. 0.66 $A^0$
• D. 0.55 $A^0$

Answer 1

Answer: (D)

0.55 $A^0$

$\lambda =\sqrt{\frac{150}{V}}{A}^{\circ }$

$=\sqrt{\frac{150}{500}}{A}^{\circ }$

$=0.55A^{\circ }$

So, the answer is option (D).