For a H atom,the energy transition from n = 1 | vedclass

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

Question

(B) The energy of an electron in a hydrogen-like species is given by $E_n = -13.6 	imes \frac{Z^2}{n^2} \ eV$. For $H$ atom $(Z=1)$,the energy differ

Vedclass · Accepted Answer

$163.2$

Vedclass · Answer

(B) The energy of an electron in a hydrogen-like species is given by $E_n = -13.6 	imes \frac{Z^2}{n^2} \ eV$. For $H$ atom $(Z=1)$,the energy difference $\Delta E = 13.6 	imes Z^2 	imes (\frac{1}{n_1^2} - \frac{1}{n_2^2}) = 13.6 	imes 1^2 	imes (\frac{1}{1^2} - \frac{1}{2^2}) = 13.6 	imes \frac{3}{4} = 10.2 \ eV$. For $Be^{3+}$ ion,the atomic number $Z = 4$. The energy difference for the same transition ($n=1$ to $n=2$) is $\Delta E' = 13.6 	imes Z^2 	imes (\frac{1}{n_1^2} - \frac{1}{n_2^2})$. $\Delta E' = 13.6 	imes 4^2 	imes (\frac{1}{1^2} - \frac{1}{2^2}) = 13.6 	imes 16 	imes \frac{3}{4} = 13.6 	imes 4 	imes 3 = 163.2 \ eV$.

For a $H$ atom,the energy transition from $n = 1$ to $n = 2$ is $10.2 \ eV$. What will be the energy for the same transition in $Be^{3+}$ ion?

Similar Questions

On the basis of Bohr's model,the radius of $3^{rd}$ orbit is

For a hydrogen atom,the energy of the $n^{th}$ orbit is given by the equation $E \propto n^y$. What is the value of $y$?

When a photon of energy $12.75 \, eV$ is completely absorbed by a hydrogen atom in its ground state,the principal quantum number of the excited state will be:

For a hydrogen atom,the energy difference between the first and second Bohr orbits is $3B/4$. What is the energy difference between the first and third Bohr orbits?

If the wavelength of light is $5.89 \times 10^{-5} \ cm$,then the energy of one photon in the given light will be ......

Vedclass Test Series

Exam Paper Generator

Online Exam Module