In a hydrogen atom,the energy difference betw | 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 $E_n = -13.6 \ eV / n^2$. The energy difference between $n = 3$ and

Vedclass · Accepted Answer

$7.2$

Vedclass · Answer

(C) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -13.6 \ eV / n^2$. The energy difference between $n = 3$ and $n = 2$ is $\Delta E = E_3 - E_2 = -13.6 \left( \frac{1}{3^2} - \frac{1}{2^2} ight) = -13.6 \left( \frac{1}{9} - \frac{1}{4} ight) = -13.6 \left( \frac{4 - 9}{36} ight) = -13.6 \left( -\frac{5}{36} ight) = \frac{13.6 	imes 5}{36} \ eV$. Given $\Delta E = E$,so $E = \frac{68}{36} \ eV = \frac{17}{9} \ eV$. The ionization energy of the $H$-atom is the energy required to move an electron from $n = 1$ to $n = \infty$,which is $E_{\infty} - E_1 = 0 - (-13.6 \ eV) = 13.6 \ eV$. Substituting $13.6 = E 	imes \frac{36}{5}$,we get $13.6 = E 	imes 7.2$. Therefore,the ionization energy is $7.2 \ E$.

In a hydrogen atom,the energy difference between the states $n = 2$ and $n = 3$ is $E \ eV$. The ionization energy of the $H$-atom is: (in $E$)

Similar Questions

An excited hydrogen atom emits light in the ultraviolet region at a frequency of $2.47 \times 10^{15} \ Hz$. Calculate the energy of a single photon. (Given: $h = 6.63 \times 10^{-34} \ J \cdot s$)

On what factors does the magnitude of the velocity of an electron in an orbit depend?

Find out the distance between the $2^{nd}$ and $3^{rd}$ orbit of the $H^{-}$ ion.

Which electron transition in a hydrogen atom requires the largest amount of energy?

In which transition of the hydrogen atom will the emitted radiation have the lowest frequency?

Vedclass Test Series

Exam Paper Generator

Online Exam Module