WBJEE · Physics · Units and Dimensions
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
- 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 & Solution
Correct Answer
(D) \(\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}\)
Step-by-step Solution
Detailed explanation
Hint: \(v=k \delta E\) \([k]=\left[\frac{v}{\delta E}\right]=\frac{\left[T^{-1}\right]}{\left[M L^{2} T^{-2}\right]}\) \([k]=\left[M^{-} L^{-2} T^{1}\right]\)
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