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

The difference in radii between the fourth and third Bohr orbits of $He^{+}$ (in $m$) is:

Which electron level would allow the hydrogen atom to absorb a photon but not emit a photon?

$A$ bulb is rated at $150 \text{ W}$, converting $8\%$ of its energy into light. If the energy of one photon is $4.42 \times 10^{-19} \text{ J}$, how many photons are emitted by the bulb per second?

Explain the photoelectric effect.

For the $H$ spectrum,the limiting line in the $Balmer$ series has a frequency of ...... $(R_\infty = 3.29 \times 10^{15} \text{ cycles/s})$

Vedclass Test Series

Exam Paper Generator

Online Exam Module