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An e- , a doubly ionized. He ion (He2+) and a proton are having the same K.E. The relation between their respective De- Broglie wavelength  \lambda _{e},\lambda _{He^{++}}\; and\; \lambda _{p}
 
Option: 1 \lambda _{e}<\lambda _{He^{++}}= \lambda _{p}
Option: 2 \lambda _{e}<\lambda _{p}<\lambda _{He^{++}}
Option: 3 \lambda _{e}>\lambda _{He^{++}} >\lambda _{p}  
Option: 4 \lambda _{e}>\lambda _{p}>\lambda _{He^{++}}

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\lambda=\frac{h}{\sqrt{2 m K}}

m_{H e}=4 m_p

\begin{array}{l} \lambda_{e}=\frac{h}{\sqrt{2 m_{e} K}} \\ \\ \lambda_{p}=\frac{h}{\sqrt{2 m_{p} K}} \end{array}

\begin{aligned} \lambda_{H_{e}} &=\frac{h}{\sqrt{2\left(m_{H e}\right) K}} =\frac{h}{\sqrt{2(4 m _p){K}}} \end{aligned}

m_{H e}>m_{p}>m_{e}

\therefore \lambda_e> \lambda_p> \lambda_{He}

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Deependra Verma

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