Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Atoms
0 votes

If stationary proton and $\;\alpha-\;$ particle dre accelerated through same potential difference , the ratio of Debrogle's wavelength will be

$(a)\;2\qquad(b)\;1\qquad(c)\;2 \sqrt{2}\qquad(d)\;None$

Can you answer this question?

1 Answer

0 votes
Answer : (c) $\;2 \sqrt{2}$
Explanation :
The gain in K.E of a charge particle of the moving through a potential difference of V is given as eV ,that is also equal to $\;\large\frac{1}{2}\;mV^2\;$ where V is velocity of charge particle . Disregarding relatiuistic effect
$\large\frac{1}{2}\;mV^2=qV \to V=\sqrt{\large\frac{2qV}{m}}$
De-Brogli's wavelength
Puting $V_{\alpha}=V_{P}\;,\large\frac{\lambda_{P}}{\lambda_{\alpha}}=\sqrt{4\times2}=2 \sqrt{2}\;.$
answered Feb 24, 2014 by yamini.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App