Question 181: 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: