In the ground state of He^{+} ion,how much en | vedclass

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

Question

(A) The energy required is the difference between the energy level of the $n=4$ state and the ground state $(n=1)$.The formula for energy difference i

Vedclass · Accepted Answer

$51$

Vedclass · Answer

(A) The energy required is the difference between the energy level of the $n=4$ state and the ground state $(n=1)$.The formula for energy difference is: $E = 13.6 	imes Z^2 	imes (\frac{1}{n_1^2} - \frac{1}{n_2^2}) \ eV$.For $He^{+}$,$Z = 2$. The ground state is $n_1 = 1$. The second line of the Balmer series corresponds to a transition from $n=4$ to $n=2$,so the electron must be excited to the $n=4$ state.$E = 13.6 	imes 2^2 	imes (\frac{1}{1^2} - \frac{1}{4^2}) \ eV$$E = 13.6 	imes 4 	imes (1 - \frac{1}{16}) \ eV$$E = 54.4 	imes (\frac{15}{16}) \ eV$$E = 51 \ eV$.

In the ground state of $He^{+}$ ion,how much energy must be supplied to the electron so that the second line of the Balmer series is obtained in its spectrum? ................. $eV$

Similar Questions

If the energy of a photon is $14 \, eV$ and it interacts with a hydrogen atom,which of the following is correct?

When electronic transition occurs from a higher energy state to a lower energy state with an energy difference equal to $\Delta E$ electron volts,the wavelength of the emitted line is approximately equal to

Henry Moseley studied the characteristic $X$-ray spectra of elements. The graph which represents his observation correctly is: (Given $\nu =$ frequency of $X$-ray emitted; $Z =$ atomic number)

The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is equal to $\frac{h^{2}}{xma_{0}^{2}}$. The value of $10x$ is ........ . ($a_{0}$ is radius of Bohr's orbit) (Nearest integer) [Given : $\pi=3.14$]

Circumference of the first orbit of hydrogen atom is given by the formula :

Vedclass Test Series

Exam Paper Generator

Online Exam Module