Question

The frequency \(v\) of the radiation emitted by an atom when an electron jumps from one orbit to another is given by \(v\) \(=\mathrm{k} \delta \mathrm{E}\), where \(\mathrm{k}\) is a constant and \(\delta \mathrm{E}\) is the change in energy level due to the transition. Then dimension of \(\mathrm{k}\) is
(A) \(\mathrm{ML}^{2} \mathrm{~T}^{-2}\)
(B) the same dimension of angular momentum
(C) \(\mathrm{ML}^{2} \mathrm{~T}^{-1}\)
(D) \(\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}\)

Answer 1

Ans:(D)
\(v=k \delta E\)
\([k]=\left[\frac{v}{\delta E}\right]=\frac{\left[T^{-1}\right]}{\left[M L^{2} T^{-2}\right]}\)
\([\mathrm{k}]=\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^{1}\right]\)

Answer 2

tadalafil 40mg generic <a href="https://ordergnonline.com/">order tadalafil 5mg pill</a> buy best erectile dysfunction pills

Answer 3

order tadalafil 40mg without prescription <a href="https://ordergnonline.com/">tadalafil usa</a> ed pills where to buy