The radius of the orbit of an electron in a H | vedclass

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

Question

(C) The radius of the orbit is given by $r_n = a_0 \frac{n^2}{Z} = 4.5 a_0$. Given orbital angular momentum $L = \frac{nh}{2\pi} = \frac{3h}{2\pi}$,so

Vedclass · Accepted Answer

$(A, C)$

Vedclass · Answer

(C) The radius of the orbit is given by $r_n = a_0 \frac{n^2}{Z} = 4.5 a_0$. Given orbital angular momentum $L = \frac{nh}{2\pi} = \frac{3h}{2\pi}$,so $n = 3$. Substituting $n=3$ into the radius formula: $4.5 = \frac{3^2}{Z} = \frac{9}{Z}$,which gives $Z = 2$. The atom is a Helium ion $(He^+)$. When the atom de-excites from $n=3$,the possible transitions are $3 \rightarrow 2$,$3 \rightarrow 1$,and $2 \rightarrow 1$. The wavelength $\lambda$ is given by $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$. For $3 \rightarrow 1$: $\frac{1}{\lambda_{3 \rightarrow 1}} = R(2^2) \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = 4R \left( 1 - \frac{1}{9} \right) = 4R \left( \frac{8}{9} \right) = \frac{32R}{9} \Rightarrow \lambda_{3 \rightarrow 1} = \frac{9}{32R}$. For $3 \rightarrow 2$: $\frac{1}{\lambda_{3 \rightarrow 2}} = R(2^2) \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = 4R \left( \frac{1}{4} - \frac{1}{9} \right) = 4R \left( \frac{5}{36} \right) = \frac{5R}{9} \Rightarrow \lambda_{3 \rightarrow 2} = \frac{9}{5R}$. For $2 \rightarrow 1$: $\frac{1}{\lambda_{2 \rightarrow 1}} = R(2^2) \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = 4R \left( 1 - \frac{1}{4} \right) = 4R \left( \frac{3}{4} \right) = 3R \Rightarrow \lambda_{2 \rightarrow 1} = \frac{1}{3R}$. The possible wavelengths are $\frac{9}{32R}$ and $\frac{9}{5R}$. Thus,the correct option is $(C)$.

Similar Questions

Let $A_n$ be the area enclosed by the $n^{th}$ orbit in a hydrogen atom. The graph of $\ln (A_n/A_1)$ against $\ln (n)$:

The magnitude of angular momentum,orbit radius,and frequency of revolution of an electron in a hydrogen atom corresponding to quantum number $n$ are $L, r$,and $f$ respectively. According to Bohr's theory of the hydrogen atom,which of the following is constant for all orbits?

An orbital electron in the ground state of hydrogen has a magnetic moment $\mu_1$. This orbital electron is excited to the $3^{rd}$ excited state by some energy transfer to the hydrogen atom. If the new magnetic moment of the electron is $\mu_2$,then:

The kinetic energy of an electron revolving around a nucleus will be

If in Bohr's atomic model,it is assumed that the force between the electron and the proton varies inversely as $r^4$,the energy of the system will be proportional to

Vedclass Test Series

Exam Paper Generator

Online Exam Module