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 λ e , λ He ++ and λ P is:

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 λ e , λ He ++ and λ P is: (A) λe<λ P <λ He + + (B) λe < λHe + +=λ P (C) λ e >λ He + +>λ P (D) λe>λ P >λ He + +

Tardigrade Admin · Accepted Answer

λ=(h/P)=(h/√2 m(K E)) λ propto (1/√ m ) ⇒ λ=( C /√ m ) m He ++ > m p > m e ∴ λ He ++<λ P <λ c

Q. An electron, a doubly ionized helium ion $(H e^{++})$ and a proton are having the same kinetic energy. The relation between their respective $d e -$ Broglie wavelengths $λ_{e}, λ_{H e^{++}}$ and $λ_{P}$ is:

$λ_{e} < λ_{P} < λ_{H e^{++}}$

$λ_{e} < λ_{H e^{++}} = λ_{P}$

$λ_{e} > λ_{H e^{++}} > λ_{P}$

$λ_{e} > λ_{P} > λ_{H e^{++}}$

$λ = \frac{h}{P} = \frac{h}{2 m ( K E )}$
$λ \propto \frac{1}{m} \Rightarrow λ = \frac{C}{m}$
$m_{H e^{++}} > m_{p} > m_{e}$
$∴ λ_{H e^{++}} < λ_{P} < λ_{c}$

Q. 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 λe​,λHe++​ and λP​ is:

λe​<λP​<λHe++​

λe​<λHe++​=λP​

λe​>λHe++​>λP​

λe​>λP​>λHe++​