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An \mathrm{\alpha } article and a carbon \mathrm{12 } atom has kinetic energy \mathrm{K }. The ratio of their de-Broglie wavelengths \mathrm{\left (\lambda _{\alpha }:\lambda _{C12} \right )} is :

Option: 1

\mathrm{1:\sqrt{3}}


Option: 2

\mathrm{\sqrt{3}:1}


Option: 3

\mathrm{3:1}


Option: 4

\mathrm{2:\sqrt{3}}


Answers (1)

best_answer

\mathrm{\lambda =\frac{h}{\sqrt{2 m k}} }\\

\mathrm{\frac{\lambda_{\alpha}}{\lambda_{c12}} =\frac{\sqrt{m_{c12}}}{\sqrt{m_{\alpha}}}=\sqrt{\frac{12}{4}}=\sqrt{3}}

Hence the correct answer is option 2

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mansi

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