If the ionisation energy for the hydrogen ato | vedclass

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

Question

(C) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula: $E_n = -\frac{13.6}{n^2} 	ext{ eV}$.For the ground st

Vedclass · Accepted Answer

$10.2$

Vedclass · Answer

(C) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula: $E_n = -\frac{13.6}{n^2} 	ext{ eV}$.For the ground state $(n=1)$, the energy is $E_1 = -\frac{13.6}{1^2} = -13.6 	ext{ eV}$.The next higher state is the first excited state $(n=2)$.The energy of the first excited state is $E_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} = -3.4 	ext{ eV}$.The energy required to excite the atom from the ground state to the first excited state is $\Delta E = E_2 - E_1$.$\Delta E = -3.4 	ext{ eV} - (-13.6 	ext{ eV}) = -3.4 	ext{ eV} + 13.6 	ext{ eV} = 10.2 	ext{ eV}$.

If the ionisation energy for the hydrogen atom is $13.6$ eV, then the energy required to excite it from the ground state to the next higher state is nearly (in $eV$)

Similar Questions

With the increase in principal quantum number,the energy difference between two successive energy levels

The diagram shows the energy levels for an electron in a certain atom. Which transition shown represents the emission of a photon with the most energy?

In a hydrogen atom,which of the following transitions will emit a photon of the highest frequency?

Let $T_{1}$ and $T_{2}$ be the energy of an electron in the first and second excited states of a hydrogen atom,respectively. According to the Bohr's model of an atom,the ratio $T_{1}: T_{2}$ is

The total energy of an electron in the $n^{th}$ stationary orbit of the hydrogen atom can be obtained by

Vedclass Test Series

Exam Paper Generator

Online Exam Module