nh/2pie. The life time of the electron in an excitated state. why electron does not merge into nucleus or lose energy while revolving around it? the wavelength of the second line of Balmer series in the hydrogen spectrum is 4861 Armstrong calculate the wavelength of the first line? Books. are denoted What is its ionisation potential? NCERT RD Sharma Cengage KC Sinha. As a result of the measurement, the electron is thrown into the corresponding eigenstate, $$\psi^{(2)}_{3/2,1/2}$$. Because the eigenstates $$\psi^{(1)}_{l,1/2;m,\pm 1/2}$$ are (presumably) orthonormal, and form a complete set, we can express the eigenstates $$\psi^{(2)}_{l,1/2;j,m_j}$$ as linear combinations of them. Maths. If the series limit of Balmer series is at 3646 A, find the wavelength of the first line of Lyman series. Legal. Because $$l=1$$ and $$s=1/2$$ can be combined together to form either $$j=3/2$$ or $$j=1/2$$ (see previously), there are also six such states: that is, $$\psi^{(2)}_{3/2,\pm 3/2}$$, $$\psi^{(2)}_{3/2,\pm 1/2}$$, and $$\psi^{(2)}_{1/2,\pm 1/2}$$. are mutually orthogonal. Physics. is 10^-8 seconds. Of course, this is because the and When the momentum of a photon is changed by an amount p’ then the corresponding change in the de-Broglie wavelength is found to be 0.20%. Of course, this is because the $$\psi^{(1)}$$ and $$\psi^{(2)}$$ eigenstates are orthonormal. Click hereto get an answer to your question ️ Determine the angular momentum of the electron of H atom in M shell. , Available for CBSE, ICSE and State Board syllabus. $\label{e11.48} \psi^{(2)}_{l-1/2,m+1/2} = \left(\frac{l-m}{2\,l+1}\right)^{1/2} \psi^{(1)}_{m,1/2} -\left(\frac{l+m+1}{2\,l+1}\right)^{1/2}\psi^{(1)}_{m+1,-1/2}.$ Here, we have neglected the common subscripts $$l,1/2$$ for the sake of clarity: that is, $$\psi^{(2)}_{l+1/2,m+1/2}\equiv \psi^{(2)}_{l,1/2;l+1/2,m+1/2}$$, et cetera. The eigenstates of , , , and , We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. The eigenstates of , , , and , Call our LearnNext Expert on 1800 419 1234 (tollfree) OR submit details below for a call back. Since the eigenstates Since can take the values , Note that the number of $$\psi^{(1)}$$ states that can appear on the right-hand side of the previous expression is limited to two by the constraint that $$m_j=m+m_s$$ [see Equation ([e11.23])], and the fact that $$m_s$$ can only take the values $$\pm 1/2$$. As an example, let us consider the $$l=1$$ states of a hydrogen atom. The angular momentum of an electron in a hydrogen atom is proportional to (where r is radius of orbit ): Offered for classes 6-12, LearnNext is a popular self-learning solution for students who strive for excellence. Have questions or comments? (See Section [sharm].) (829)-(837) is neatly summed Find the velocity and moment of the scattered electron. Once $$x$$ is specified, Equations ([e11.30]) and ([e11.45]) can be solved for $$\alpha$$ and $$\beta$$. Radius of the nth orbit rn ∝ n2/Z      ⇒    n ∝ (rn)½ Hence, the simultaneous eigenstates of $$L^2$$, $$S^{\,2}$$, $$L_z$$, and $$S_z$$ can be written in the separable form $\label{e11.28} \psi^{(1)}_{l,1/2;m,\pm 1/2} = Y_{l,m}\,\chi_\pm.$ Here, it is understood that orbital angular momentum operators act on the spherical harmonic functions, $$Y_{l,m}$$, whereas spin angular momentum operators act on the spinors, $$\chi_\pm$$. For instance, It follows from Eqs. Assuming that the $$\psi^{(2)}$$ eigenstates are properly normalized, we have, $\label{e11.30} \alpha^{\,2} + \beta^{\,2} = 1.$, Now, it follows from Equation ([e11.26]) that, $\label{e11.31} J^{\,2}\,\psi^{(2)}_{l,1/2;j,m+1/2}= j\,(j+1)\,\hbar^{\,2}\,\psi^{(2)}_{l,1/2;j,m+1/2},$ where [see Equation ([e11.12])] $\label{e11.32} J^{\,2} = L^2+S^{\,2} +2\,L_z\,S_z+ L_+\,S_-+L_-\,S_+.$ Moreover, according to Equations ([e11.28]) and ([e11.29]), we can write $\label{e11.33} \psi^{(2)}_{l,1/2;j,m+1/2} = \alpha\,Y_{l,m}\,\chi_+ + \beta\, Y_{l,m+1}\,\chi_-.$ Recall, from Equations ([eraise]) and ([elow]), that \begin{aligned} \label{e11.34} L_+\,Y_{l,m} &= [l\,(l+1)-m\,(m+1)]^{1/2}\,\hbar\,Y_{l,m+1},\\[0.5ex] L_-\,Y_{l,m} &= [l\,(l+1)-m\,(m-1)]^{1/2}\,\hbar\,Y_{l,m-1}.\end{aligned} By analogy, when the spin raising and lowering operators, $$S_\pm$$, act on a general spinor, $$\chi_{s,m_s}$$, we obtain \begin{aligned} S_+\,\chi_{s,m_s} &= [s\,(s+1)-m_s\,(m_s+1)]^{1/2}\,\hbar\,\chi_{s,m_s+1},\\[0.5ex] S_-\,\chi_{s,m_s} &= [s\,(s+1)-m_s\,(m_s-1)]^{1/2}\,\hbar\,\chi_{s,m_s-1}.\end{aligned} For the special case of spin one-half spinors (i.e., $$s=1/2, m_s=\pm 1/2$$), the previous expressions reduce to $S_+\,\chi_+=S_-\,\chi_- = 0,$ and $\label{e11.39} S_\pm\,\chi_\mp = \hbar\,\chi_\pm.$, It follows from Equations ([e11.32]) and ([e11.34])–([e11.39]) that \begin{aligned} J^{\,2}\,Y_{l,m}\,\chi_+&= [l\,(l+1)+3/4+m]\,\hbar^{\,2}\,Y_{l,m}\,\chi_+\nonumber\\[0.5ex] &\phantom{=}+ [l\,(l+1)-m\,(m+1)]^{1/2}\,\hbar^{\,2}\,Y_{l,m+1}\,\chi_-,\end{aligned} and \begin{aligned} J^{\,2}\,Y_{l,m+1}\,\chi_-&= [l\,(l+1)+3/4-m-1]\,\hbar^{\,2}\,Y_{l,m+1}\,\chi_-\nonumber\\[0.5ex] &\phantom{=}+ [l\,(l+1)-m\,(m+1)]^{1/2}\,\hbar^{\,2}\,Y_{l,m}\,\chi_+.\end{aligned} Hence, Equations ([e11.31]) and ([e11.33]) yield, \begin{aligned} \label{e11.42} (x - m)\,\alpha - [l\,(l+1)-m\,(m+1)]^{1/2}\,\beta &= 0,\\[0.5ex] -[l\,(l+1)-m\,(m+1)]^{1/2}\,\alpha +(x+m+1)\,\beta&= 0,\label{e11.43}\end{aligned} where $x = j\,(j+1) - l\,(l+1) - 3/4.$ Equations ([e11.42]) and ([e11.43]) can be solved to give $x\,(x+1) = l\,(l+1),$ and, $\label{e11.45} \frac{\alpha}{\beta} = \frac{[(l-m)\,(l+m+1)]^{1/2}}{x-m}.$ It follows that $$x=l$$ or $$x=-l-1$$, which corresponds to $$j=l+1/2$$ or $$j=l-1/2$$, respectively.

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