JEE MAIN - Physics (2020 - 6th September Morning Slot - No. 19)
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 :
$$\lambda $$e > $$\lambda $$He++ > $$\lambda $$p
$$\lambda $$e < $$\lambda $$p < $$\lambda $$He++
$$\lambda $$e > $$\lambda $$p > $$\lambda $$He++
$$\lambda $$e < $$\lambda $$He++ = $$\lambda $$p
Explanation
$$\lambda $$ = $${h \over P}$$ = $${h \over {\sqrt {2m\left( {KE} \right)} }}$$
$$ \therefore $$ $$\lambda $$ $$ \propto $$ $${1 \over {\sqrt m }}$$
mHe++ > mp > me
$$ \therefore $$ $$\lambda $$e > $$\lambda $$p > $$\lambda $$He++
$$ \therefore $$ $$\lambda $$ $$ \propto $$ $${1 \over {\sqrt m }}$$
mHe++ > mp > me
$$ \therefore $$ $$\lambda $$e > $$\lambda $$p > $$\lambda $$He++
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