The ionization energy of a hydrogen atom is 1 | vedclass

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

Question

(D) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} \ eV$.For the ground state,$n = 1$.The sec

Vedclass · Accepted Answer

$1.5$

Vedclass · Answer

(D) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} \ eV$.For the ground state,$n = 1$.The second excited state corresponds to $n = 3$.The energy of the electron in the second excited state is $E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} \approx -1.51 \ eV$.The energy required to ionize the atom from this state is the energy needed to bring the electron to $n = \infty$,which is $E_{\infty} - E_3 = 0 - (-1.51) = 1.51 \ eV$.

The ionization energy of a hydrogen atom is $13.6 \ eV$. The energy required to ionize a hydrogen atom in its second excited state is ........... $eV$.

Similar Questions

Calculate the energy in joule corresponding to light of wavelength $45 \ nm$. (Planck's constant,$h = 6.63 \times 10^{-34} \ J \ s$,speed of light,$c = 3 \times 10^8 \ m \ s^{-1}$)

The ionization energy of an electron in an excited state of a hydrogen atom is $+0.85 \, eV$. What will be the energy of the emitted photon in $eV$ when it returns to the ground state?

The value of Rydberg constant $\left(R_H\right)$ is $2.18 \times 10^{-18} \ J$. The velocity of an electron having mass $9.1 \times 10^{-31} \ kg$ in Bohr's first orbit of a hydrogen atom $= \dots \dots \dots \times 10^5 \ ms^{-1}$ (nearest integer).

The angular momentum of an electron in a given orbit is $J$. Its kinetic energy will be:

Which of the following electron transitions in a hydrogen atom will require the largest amount of energy?

Vedclass Test Series

Exam Paper Generator

Online Exam Module