If the total energy of an electron in a hydro | vedclass

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

Question

(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} 	ext{ eV}$. Given $E_n = -3.4 	ext{ eV}$

Vedclass · Accepted Answer

$6.6 	imes 10^{-10} 	ext{ m}$

Vedclass · Answer

(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} 	ext{ eV}$. Given $E_n = -3.4 	ext{ eV}$,we have $-\frac{13.6}{n^2} = -3.4$,which gives $n^2 = 4$,so $n = 2$. According to Bohr's postulate,$mvr = \frac{nh}{2\pi}$,so the de-Broglie wavelength $\lambda = \frac{h}{mv} = \frac{2\pi r}{n}$. For hydrogen,the radius of the $n^{th}$ orbit is $r_n = 0.529 	imes n^2 	ext{ Å}$. For $n = 2$,$r_2 = 0.529 	imes 2^2 = 2.116 	ext{ Å}$. Substituting these values,$\lambda = \frac{2 	imes \pi 	imes 2.116}{2} = 6.64 	ext{ Å} = 6.64 	imes 10^{-10} 	ext{ m}$.

If the total energy of an electron in a hydrogen atom in an excited state is $-3.4 \text{ eV}$,then the de-Broglie wavelength of the electron is

Similar Questions

State the de-Broglie principle and explain it.

What will be the de-Broglie wavelength of an electron moving with a velocity of $1.2 \times 10^5 \ m \ s^{-1}$?

Which one of the following explains light both as a stream of particles and as wave motion?

If $\lambda _L, \lambda _M$ and $\lambda _N$ are the wavelengths of an electron in $L, M, N$ energy levels of $H^{-}$ atom respectively,what is their decreasing order?

$A$ cricket ball of $0.5 \ kg$ is moving with a velocity of $100 \ m/s$. The wavelength associated with its motion is

Vedclass Test Series

Exam Paper Generator

Online Exam Module