A) \[\frac{E}{E}=-\frac{2Q}{Q}\]
B) \[\Rightarrow \]
C) \[E=-\frac{E}{2}\]
D) \[3.9\Omega \]
Correct Answer: D
Solution :
The de-Broglie wavelength is given by \[\frac{(n+1)\lambda }{2}\] where h is Plancks constant and p is momentum. Also \[n(\lambda +1)\] where m is mass, u is velocity \[n\lambda \] Putting the numerical values, we have \[[M{{L}^{3}}{{T}^{-3}}]\] \[[M{{L}^{-1}}{{T}^{-1}}]\] \[[M{{L}^{2}}{{T}^{-2}}]\] \[[M{{T}^{-2}}]\] \[\sqrt{\frac{1}{{{\varepsilon }_{0}}{{\mu }_{0}}}}\]
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