Search any question & find its solution
Question:
Answered & Verified by Expert
An electron makes a transition from an excited state to the ground state of a hydrogen like atom. Out of the following statements which one is correct?
Options:
Solution:
2205 Upvotes
Verified Answer
The correct answer is:
Kinetic energy increases but potential energy and total energy decreases
$$
\begin{aligned}
& \text { Potential energy }=-\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{ze}^2}{\mathrm{r}} \\
& \text { Kinetic energy }=-\frac{1}{2}(\text { P.E. })=\frac{1}{8 \pi \varepsilon_0} \cdot \frac{\mathrm{ze}^2}{\mathrm{r}} \\
& \text { Total energy } \left.\frac{1}{2} \text { (P.E. }\right)=-\frac{1}{8 \pi \varepsilon_0} \frac{\mathrm{ze}^2}{\mathrm{r}}
\end{aligned}
$$
As $r$ decreases K.E, increases but P.E. and T.E. decreases (since they are negative).
\begin{aligned}
& \text { Potential energy }=-\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{ze}^2}{\mathrm{r}} \\
& \text { Kinetic energy }=-\frac{1}{2}(\text { P.E. })=\frac{1}{8 \pi \varepsilon_0} \cdot \frac{\mathrm{ze}^2}{\mathrm{r}} \\
& \text { Total energy } \left.\frac{1}{2} \text { (P.E. }\right)=-\frac{1}{8 \pi \varepsilon_0} \frac{\mathrm{ze}^2}{\mathrm{r}}
\end{aligned}
$$
As $r$ decreases K.E, increases but P.E. and T.E. decreases (since they are negative).
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.