From the following energy levels of a hydroge | 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 orbital of a hydrogen atom is given by the formula: $E_n = -2.18 	imes 10^{-18} \cdot \frac{Z^2}{n^2} \ J$

Vedclass · Accepted Answer

$0, -0.242 	imes 10^{-18}$

Vedclass · Answer

(C) The energy of an electron in the $n$th orbital of a hydrogen atom is given by the formula: $E_n = -2.18 	imes 10^{-18} \cdot \frac{Z^2}{n^2} \ J$. For a hydrogen atom, $Z = 1$. Thus, $E_n = -\frac{2.18 	imes 10^{-18}}{n^2} \ J$. For $n = \infty$, $E_{\infty} = -\frac{2.18 	imes 10^{-18}}{\infty} = 0 \ J$. For $n = 3$, $E_3 = -\frac{2.18 	imes 10^{-18}}{3^2} = -\frac{2.18 	imes 10^{-18}}{9} = -0.242 	imes 10^{-18} \ J$. Therefore, $E_{\infty} = 0 \ J$ and $E_3 = -0.242 	imes 10^{-18} \ J$. Hence, option $(c)$ is the correct answer.

From the following energy levels of a hydrogen atom, the values of $E_{\infty}$ and $E_3$ in $J$ are, respectively: $E_{\infty} = \ldots \ldots \ldots$ $E_3 = \ldots \ldots \ldots$ Given: $E_2 = -0.545 \times 10^{-18} \ J$, $E_1 = -2.18 \times 10^{-18} \ J$.

Similar Questions

The energy of a photon of wavelength $(\lambda) = 1 \, m$ is (Planck's constant $= 6.626 \times 10^{-34} \, J \, s$,speed of light $= 3 \times 10^8 \, ms^{-1}$)

The ionization energy of sodium in $kJ \ mol^{-1}$,if electromagnetic radiation of wavelength $242 \ nm$ is just sufficient to ionize a sodium atom,is . . . . . . .

The radius of the first excited state of a Helium ion $(He^+)$ is given by (where $a_0$ is the radius of the first stationary state of the hydrogen atom):

Explain Bohr's atomic model.

When $\alpha$-particles are sent through a thin metal foil,most of them go straight through the foil because (one or more are correct)

Vedclass Test Series

Exam Paper Generator

Online Exam Module