Question 160: Dual Nature of Matter and Radiation

Question

An electron, a doubly ionized helium ion (He
++
) and a proton are having the same kinetic energy.
The relation between their respective de-Broglie wavelengths $$\lambda $$
e
, $$\lambda $$
He
++
and $$\lambda $$
p
is :

Accepted Answer

$$\lambda $$ = $${h \over P}$$ = $${h \over {\sqrt {2m\left( {KE} ight)} }}$$
$$ 	herefore $$ $$\lambda $$ $$ \propto $$ $${1 \over {\sqrt m }}$$
m
He
++
> m
p
> m
e
$$ 	herefore $$ $$\lambda $$
e
> $$\lambda $$
p
> $$\lambda $$
He
++

Answer

$$\lambda $$
e
> $$\lambda $$
He
++
> $$\lambda $$
p

Answer

$$\lambda $$
e
< $$\lambda $$
p
< $$\lambda $$
He
++

Answer

$$\lambda $$
e
> $$\lambda $$
p
> $$\lambda $$
He
++

Answer

$$\lambda $$
e
< $$\lambda $$
He
++
= $$\lambda $$
p

Dual Nature of Matter and Radiation

An electron, a doubly ionized helium ion (He ++ ) and a proton are having the same kinetic energy. The relation between their respective de-Broglie wavelengths $\lambda e , \lambda He ++ and \lambda $ p is :

Choose the correct answer: