Search any question & find its solution
Question:
Answered & Verified by Expert
The frequency $v$ of the radiation emitted by an atom when an electron jumps from one orbit to another is given by $=\mathrm{k} \delta \mathrm{E}$, 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
Options:
Solution:
1237 Upvotes
Verified Answer
The correct answer is:
$\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}$
Hint:
$v=k \delta E$
$[k]=\left[\frac{v}{\delta E}\right]=\frac{\left[T^{-1}\right]}{\left[M L^{2} T^{-2}\right]}$
$[k]=\left[M^{-} L^{-2} T^{1}\right]$
$v=k \delta E$
$[k]=\left[\frac{v}{\delta E}\right]=\frac{\left[T^{-1}\right]}{\left[M L^{2} T^{-2}\right]}$
$[k]=\left[M^{-} L^{-2} T^{1}\right]$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.