In the Bohr's model , for unielectronic species following symbols are used
`r_(n,z)to` Radius of `n^"th"` orbit with atomic number "z"
`U_(n,z)to` Potential energy of electron in `n^"th"` orbit with atomic number "z"
`K_(n,z)to` Kinetic energy of electron in `n^"th"` orbit with atomic number "z"
`V_(n,z)to` Velocity of electon in `n^"th"` orbit with atomic number "z"
`T_(n,z)to` Time period of revolution of electon in `n^"th"` orbit with atomic number "z"
Calculate z in all in cases.
(i)`U_(1,2):K_(1,z)=-8:9`
(ii)`r_(1,z):r_(2,1) =1:12`
(iii)`v_(1,z):v_(3,1)=15:1`
(iv)`T_(1,2):T_(2,z)=9:32`
Report your answer as (2r-p-q-s) where p,q,r and s represents the value of "z" in parts (i),(ii),(iii),(iv).

Answer

To solve the problem, we will analyze each part (i) to (iv) step by step using the Bohr model equations for unielectronic species.

### Given Information:
1. **Bohr's Model Equations:**
- Radius of nth orbit: \( r_{n,z} = \frac{n^2}{z} \cdot r_0 \) (where \( r_0 \) is a constant)
- Potential energy: \( U_{n,z} = -\frac{z^2 e^4 m}{2 \hbar^2 n^2} \)
- Kinetic energy: \( K_{n,z} = \frac{z^2 e^4 m}{8 \hbar^2 n^2} \)
- Velocity: \( V_{n,z} = \frac{z e^2}{2 \hbar} \cdot \frac{1}{n} \)
- Time period: \( T_{n,z} = \frac{2 \pi n}{V_{n,z}} \)

### Part (i): \( U_{1,2}: K_{1,z} = -8:9 \)

1. **Calculate \( U_{1,2} \) and \( K_{1,z} \):**
- \( U_{1,2} = -\frac{2^2 e^4 m}{2 \hbar^2} = -\frac{4 e^4 m}{2 \hbar^2} = -\frac{2 e^4 m}{\hbar^2} \)
- \( K_{1,z} = \frac{z^2 e^4 m}{8 \hbar^2} \)

2. **Set up the ratio:**
\[ \frac{U_{1,2}}{K_{1,z}} = \frac{-\frac{2 e^4 m}{\hbar^2}}{\frac{z^2 e^4 m}{8 \hbar^2}} = -\frac{16}{z^2} \]
Given \( -\frac{16}{z^2} = -\frac{8}{9} \)

3. **Solve for \( z \):**
\[ \frac{16}{z^2} = \frac{8}{9} \implies 16 \cdot 9 = 8 z^2 \implies 144 = 8 z^2 \implies z^2 = 18 \implies z = 3 \]

### Part (ii): \( r_{1,z}: r_{2,1} = 1:12 \)

1. **Calculate \( r_{1,z} \) and \( r_{2,1} \):**
- \( r_{1,z} = \frac{1^2}{z} \cdot r_0 = \frac{r_0}{z} \)
- \( r_{2,1} = \frac{2^2}{1} \cdot r_0 = 4r_0 \)

2. **Set up the ratio:**
\[ \frac{r_{1,z}}{r_{2,1}} = \frac{\frac{r_0}{z}}{4r_0} = \frac{1}{4z} \]
Given \( \frac{1}{4z} = \frac{1}{12} \)

3. **Solve for \( z \):**
\[ 12 = 4z \implies z = 3 \]

### Part (iii): \( v_{1,z}: v_{3,1} = 15:1 \)

1. **Calculate \( v_{1,z} \) and \( v_{3,1} \):**
- \( v_{1,z} = \frac{z e^2}{2 \hbar} \)
- \( v_{3,1} = \frac{1 e^2}{6 \hbar} \)

2. **Set up the ratio:**
\[ \frac{v_{1,z}}{v_{3,1}} = \frac{\frac{z e^2}{2 \hbar}}{\frac{e^2}{6 \hbar}} = \frac{6z}{2} = 3z \]
Given \( 3z = 15 \)

3. **Solve for \( z \):**
\[ z = 5 \]

### Part (iv): \( T_{1,2}: T_{2,z} = 9:32 \)

1. **Calculate \( T_{1,2} \) and \( T_{2,z} \):**
- \( T_{1,2} = \frac{2 \pi \cdot 1}{\frac{2 e^2}{2 \hbar}} = \frac{2 \pi \cdot 2 \hbar}{e^2} \)
- \( T_{2,z} = \frac{2 \pi \cdot 2}{\frac{z e^2}{2 \hbar}} = \frac{4 \pi \cdot 2 \hbar}{z e^2} \)

2. **Set up the ratio:**
\[ \frac{T_{1,2}}{T_{2,z}} = \frac{\frac{4 \pi \hbar}{e^2}}{\frac{4 \pi \cdot 2 \hbar}{z e^2}} = \frac{z}{2} \]
Given \( \frac{z}{2} = \frac{9}{32} \)

3. **Solve for \( z \):**
\[ z = \frac{9 \cdot 2}{32} = \frac{18}{32} = \frac{9}{16} \]

### Final Answers:
- (i) \( z = 3 \)
- (ii) \( z = 3 \)
- (iii) \( z = 5 \)
- (iv) \( z = \frac{9}{16} \)

### Reporting the Answer:
The final answer is \( (2r - p - q - s) \) where \( p = 3, q = 3, r = 5, s = \frac{9}{16} \).