JEE MAIN - Chemistry (2016 (Offline) - No. 18)
A stream of electrons from a heated filament was passed between two charged plates kept at a potential
difference V esu. If e and m are charge and mass of an electron, respectively, then the value of h / $$\lambda $$ (where
$$\lambda $$ is wavelength associated with electron wave) is given by :
2meV
$$\sqrt {meV} $$
$$\sqrt {2meV} $$
meV
Explanation
We know, Kinetic Energy (KE) = $${1 \over 2}m{v^2}$$
$$\therefore$$ mv2 = 2KE
$$ \Rightarrow $$ m2v2 = 2mKE
$$ \Rightarrow $$ mv = $$\sqrt {2m{K_E}} $$
For an electron of charge 'e' which is passes through 'V' volt, kinetic energy of electron will be
KE = eV
$$\therefore$$ m = $$\sqrt {2meV} $$
We know de-Broglie wavelength for an electron$$\left( \lambda \right)$$ = $${h \over p} = {h \over {mv}}$$
$$\therefore$$ $${h \over \lambda }$$ = mv = $$\sqrt {2meV} $$
$$\therefore$$ mv2 = 2KE
$$ \Rightarrow $$ m2v2 = 2mKE
$$ \Rightarrow $$ mv = $$\sqrt {2m{K_E}} $$
For an electron of charge 'e' which is passes through 'V' volt, kinetic energy of electron will be
KE = eV
$$\therefore$$ m = $$\sqrt {2meV} $$
We know de-Broglie wavelength for an electron$$\left( \lambda \right)$$ = $${h \over p} = {h \over {mv}}$$
$$\therefore$$ $${h \over \lambda }$$ = mv = $$\sqrt {2meV} $$
Comments (0)
