InterviewSolution
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InterviewSolution
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A proton, a neutron, an electron and an `alpha`-particle have same energy. Then their de-Broglie wavelengths compare asA. `lamda_(p)=lamda_(n)gtlamda_(e)gtlamda_(alpha)`B. `lamda_(alpha)ltlamda_(p)=lamda_(n)ltlamda_(e)`C. `lamda_(e)ltlamda_(p)=lamda_(n)gtlamda_(alpha)`D. `lamda_(e)=lamda_(p)=lamda_(n)=lamda_(alpha)` |
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Answer» Correct Answer - B (b) : Kinetic energy of particle, `K=(1)/(2)mv^(2)` or `mv=sqrt(2mK)` de Bloglie wavelength, `lamda=(h)/(mv)=(h)/(sqrt2mK)` `"For the given value of K",lamdaprop(1)/(sqrt(2mK))` `:.lamda_(p):lamda_(n):lamda_(e):lamda_(alpha)=(1)/(sqrt(m_(p))):(1)/(sqrt(m_(n))):(1)/(sqrt(m_(e))):(1)/(sqrt(m_(alpha)))` `"Since "m_(p)=m_(n),"hence "lamda_(p)=lamda_(n)` `"As "m_(alpha)gtm_(p),"therefore"lamda_(alpha)ltlamda_(p)` `"As " m_(e)ltm_(n),"therefore"lamda_(e)gtlamda_(n)` `"Hence "lamda_(alpha)ltlamda_(p)=lamda_(n)ltlamda_(e)` |
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