An energy of 24.6 eV is required to remove on | vedclass

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

Question

(A) The energy required to remove the first electron from a neutral helium atom $(He)$ is given as $E_1 = 24.6\ eV$.After removing the first electron,

Vedclass · Accepted Answer

$79$

Vedclass · Answer

(A) The energy required to remove the first electron from a neutral helium atom $(He)$ is given as $E_1 = 24.6\ eV$.After removing the first electron,the remaining system is a $He^+$ ion,which is a hydrogen-like atom with atomic number $Z = 2$.The energy required to remove the second electron from the ground state of a hydrogen-like ion is given by the formula $E_n = 13.6 	imes Z^2\,eV$.For $He^+$,$Z = 2$,so the energy required to remove the second electron is $E_2 = 13.6 	imes (2)^2 = 13.6 	imes 4 = 54.4\ eV$.The total energy required to remove both electrons is the sum of the energies required for each step: $E_{total} = E_1 + E_2 = 24.6\ eV + 54.4\ eV = 79\ eV$.

An energy of $24.6\ eV$ is required to remove one of the electrons from a neutral helium atom. The energy (in $eV$) required to remove both the electrons from a neutral helium atom is

Similar Questions

The nature of the sun's spectrum is

In a $He-Ne$ laser,the metastable state exists in:

The graph between the square root of the frequency of a specific line of characteristic spectrum of $X$-rays and the atomic number of the target will be

$A$ neutron collides head-on with a stationary hydrogen atom in its ground state.

$A$ neutron moving with a speed $v$ makes a head-on collision with a stationary hydrogen atom in its ground state. The minimum kinetic energy of the neutron for which an inelastic collision will take place is....$eV$

Vedclass Test Series

Exam Paper Generator

Online Exam Module