For a hydrogen atom, the ratio of the largest wavelength of Lyman series to that of the Balmer series is.

### Related Formula textKE = -E$\text{KE} = -E$ textPE = 2E$\text{PE} = 2E$ textKE = frac12 |textPE|$$\text{KE} = \frac{1}{2} |\text{PE}|$$ ### Core Logic In any allowed Bohr orbit (for any value of principal quantum number n$n$), the relationship between kinetic energy (KE), potential energy (PE), and total energy (E) strictly obeys the virial theorem for a Coulombic force field. |textPE| = 2 times textKE$$|\text{PE}| = 2 \times \text{KE}$$ ### Step 1: Formulate the Ratio The question asks for the ratio of the magnitude of KE to the magnitude of PE. fractextKE|textPE| = frac12$$\frac{\text{KE}}{|\text{PE}|} = \frac{1}{2}$$ This ratio is independent of the orbit state. Even though it is the 5^textth$5^{\text{th}}$ excited state, the ratio remains frac12$\frac{1}{2}$. ### Pattern Recognition Energy relationships in Bohr orbits (and planetary motion): K = -E = -U/2$K = -E = -U/2$. The state number (n$n$) is given solely as a distractor. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Atoms

### Related Formula E_n = -frac13.6n^2 mathrm~eV$$E_n = -\frac{13.6}{n^2} \mathrm{~eV}$$ textNumber of transitions = fracn(n - 1)2$$\text{Number of transitions} = \frac{n(n - 1)}{2}$$ ### Core Logic First, identify the principal quantum number n$n$ corresponding to the energy -0.85 mathrm~eV$-0.85 \mathrm{~eV}$. Then, use the combinatorics formula to find the total possible downward emission transitions. ### Step 1: Find Quantum State E_n = -frac13.6n^2$$E_n = -\frac{13.6}{n^2}$$ -0.85 = -frac13.6n^2$$-0.85 = -\frac{13.6}{n^2}$$ n^2 = frac13.60.85 = 16$$n^2 = \frac{13.6}{0.85} = 16$$ n = 4$n = 4$ ### Step 2: Calculate Transitions Maximum number of transitions from n=4$n=4$ to lower levels (n=3, 2, 1$n=3, 2, 1$): = fracn(n - 1)2$$= \frac{n(n - 1)}{2}$$ = frac4(4 - 1)2 = frac122 = 6$$= \frac{4(4 - 1)}{2} = \frac{12}{2} = 6$$ ### Pattern Recognition Energy states in Hydrogen are heavily standardized: n=1 rightarrow -13.6$n=1 \rightarrow -13.6$, n=2 rightarrow -3.4$n=2 \rightarrow -3.4$, n=3 rightarrow -1.51$n=3 \rightarrow -1.51$, n=4 rightarrow -0.85$n=4 \rightarrow -0.85$. Recognize -0.85 mathrm~eV$-0.85 \mathrm{~eV}$ as state 4 immediately. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Atoms

### Related Formula frac1lambda = R Z^2 left[ frac1n_1^2 - frac1n_2^2 right]$$\frac{1}{\lambda} = R Z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$$ ### Core Logic For the first member of the Lyman series of hydrogen (n_1 = 1, n_2 = 2$n_1 = 1, n_2 = 2$): frac1lambda = frac13.6 Z^2hc left[ frac11^2 - frac12^2 right]$$\frac{1}{\lambda} = \frac{13.6 Z^2}{hc} \left[ \frac{1}{1^2} - \frac{1}{2^2} \right]$$ frac1lambda = frac13.6 Z^2hc left[ frac34 right] dots dots (texti)$$\frac{1}{\lambda} = \frac{13.6 Z^2}{hc} \left[ \frac{3}{4} \right] \dots \dots (\text{i})$$ ### Step 2: Second Member Calculation For the second member of the Lyman series (n_1 = 1, n_2 = 3$n_1 = 1, n_2 = 3$): frac1lambda' = frac13.6 Z^2hc left[ frac11^2 - frac13^2 right]$$\frac{1}{\lambda'} = \frac{13.6 Z^2}{hc} \left[ \frac{1}{1^2} - \frac{1}{3^2} \right]$$ frac1lambda' = frac13.6 Z^2hc left[ frac89 right] dots dots (textii)$$\frac{1}{\lambda'} = \frac{13.6 Z^2}{hc} \left[ \frac{8}{9} \right] \dots \dots (\text{ii})$$ ### Step 3: Ratio On dividing equation (i) by (ii): fraclambda'lambda = frac3/48/9 = frac34 times frac98$$\frac{\lambda'}{\lambda} = \frac{3/4}{8/9} = \frac{3}{4} \times \frac{9}{8}$$ fraclambda'lambda = frac2732$$\frac{\lambda'}{\lambda} = \frac{27}{32}$$ lambda' = frac2732 lambda$$\lambda' = \frac{27}{32} \lambda$$ ### Pattern Recognition Rydberg ratios between members of the same series are purely derived from the bracket terms [1/n_1^2 - 1/n_2^2]$[1/n_1^2 - 1/n_2^2]$. For Lyman 1st and 2nd, the ratio is (3/4) / (8/9) = 27/32$(3/4) / (8/9) = 27/32$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Atoms