The energy required to excite an electron fro | vedclass

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

Question

(A) The energy difference for an electron transition in a hydrogen-like atom is given by the formula: $\Delta E = 13.6 	imes Z^2 	imes (\frac{1}{n_1

Vedclass · Accepted Answer

$12.1$

Vedclass · Answer

(A) The energy difference for an electron transition in a hydrogen-like atom is given by the formula: $\Delta E = 13.6 	imes Z^2 	imes (\frac{1}{n_1^2} - \frac{1}{n_2^2}) \ eV$.For a hydrogen atom,$Z = 1$,$n_1 = 1$,and $n_2 = 3$.Substituting these values:$\Delta E = 13.6 	imes 1^2 	imes (\frac{1}{1^2} - \frac{1}{3^2}) \ eV$.$\Delta E = 13.6 	imes (1 - \frac{1}{9}) \ eV$.$\Delta E = 13.6 	imes \frac{8}{9} \ eV$.$\Delta E = 12.088 \approx 12.1 \ eV$.

The energy required to excite an electron from the first orbit to the third orbit in a hydrogen atom is .......... $eV$.

Similar Questions

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

The potential energy of an electron in the hydrogen atom is $-6.8 \ eV$. Indicate in which excited state the electron is present.

Angular momentum of an electron in the second shell of a $Li^{+2}$ ion is:

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?

Binding energy of a hydrogen atom is $13.6 \ eV$. The binding energy of a singly ionized helium atom is : ............ $eV$

Vedclass Test Series

Exam Paper Generator

Online Exam Module