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: