Based on the equation E = -2.178 times 10^{-1 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The energy of an electron at an infinite distance from the nucleus is defined as zero. As an electron approaches the nucleus,the electrostatic att

Vedclass · Accepted Answer

For $n = 1$,the electron has a more negative energy than it does for $n = 6$,which means that the electron is more loosely bound in the smallest allowed orbit.

Vedclass · Answer

(C) The energy of an electron at an infinite distance from the nucleus is defined as zero. As an electron approaches the nucleus,the electrostatic attraction increases,causing the energy of the electron to decrease and become negative. Therefore,as the value of $n$ decreases (i.e.,the orbit is closer to the nucleus),the energy of the electron becomes more negative,meaning it is more tightly bound,not loosely bound. Thus,the statement in option $C$ is incorrect because it claims the electron is more loosely bound in the smallest orbit.

Based on the equation $E = -2.178 \times 10^{-18} \ J \left( \frac{Z^2}{n^2} \right)$,certain conclusions are written. Which of them is not correct?

Similar Questions

Which of the following relations is correct for the energy of the ground state of hydrogen-like species?

When $\alpha$-particles are sent through a thin metal foil,most of them go straight through the foil because (one or more are correct)

$A$ quantum will have more energy if :

Vedclass Test Series

Exam Paper Generator

Online Exam Module