Consider three electron jumps described below for the hydrogen atom
`{:["X: n=3 to n=1"],["Y: n=4 to n=2"],["Z: n=5 to n=3"]:}`
(a) The photon emitted in which transition ` X, Y` or (Z) will have shortest wavelength ?
(b) For which transition will the electron experience the longest charge in orbit radius ?

Explanation

To solve the given question regarding electron transitions in a hydrogen atom, we will follow these steps:

### Step 1: Understand the transitions
We have three transitions:
- **X**: n = 3 to n = 1
- **Y**: n = 4 to n = 2
- **Z**: n = 5 to n = 3

### Step 2: Calculate the energy changes for each transition
The energy of an electron in a hydrogen atom can be calculated using the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]

#### For Transition X (n = 3 to n = 1):
1. Calculate \(E_3\):
\[ E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} \approx -1.51 \, \text{eV} \]
2. Calculate \(E_1\):
\[ E_1 = -\frac{13.6}{1^2} = -13.6 \, \text{eV} \]
3. Change in energy (\(\Delta E_X\)):
\[ \Delta E_X = E_1 - E_3 = -13.6 - (-1.51) = -12.09 \, \text{eV} \]

#### For Transition Y (n = 4 to n = 2):
1. Calculate \(E_4\):
\[ E_4 = -\frac{13.6}{4^2} = -\frac{13.6}{16} \approx -0.85 \, \text{eV} \]
2. Calculate \(E_2\):
\[ E_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} \approx -3.4 \, \text{eV} \]
3. Change in energy (\(\Delta E_Y\)):
\[ \Delta E_Y = E_2 - E_4 = -3.4 - (-0.85) = -2.55 \, \text{eV} \]

#### For Transition Z (n = 5 to n = 3):
1. Calculate \(E_5\):
\[ E_5 = -\frac{13.6}{5^2} = -\frac{13.6}{25} \approx -0.544 \, \text{eV} \]
2. Calculate \(E_3\) (already calculated):
\[ E_3 \approx -1.51 \, \text{eV} \]
3. Change in energy (\(\Delta E_Z\)):
\[ \Delta E_Z = E_3 - E_5 = -1.51 - (-0.544) = -0.966 \, \text{eV} \]

### Step 3: Determine which transition has the shortest wavelength
The energy of the emitted photon is related to its wavelength by:
\[ E = \frac{hc}{\lambda} \]
Where \(h\) is Planck's constant and \(c\) is the speed of light. Higher energy corresponds to shorter wavelength.

From our calculations:
- \(\Delta E_X \approx -12.09 \, \text{eV}\)
- \(\Delta E_Y \approx -2.55 \, \text{eV}\)
- \(\Delta E_Z \approx -0.966 \, \text{eV}\)

Since \(\Delta E_X\) has the highest magnitude, transition **X** (n = 3 to n = 1) will emit a photon with the shortest wavelength.

### Step 4: Determine which transition has the longest change in orbit radius
The radius of the electron in a hydrogen atom is given by:
\[ r_n = 0.529 \times n^2 \]

#### Calculate the change in radius for each transition:
1. **For X (n = 3 to n = 1)**:
\[ \Delta r_X = r_1 - r_3 = 0.529 \times (1^2 - 3^2) = 0.529 \times (1 - 9) = 0.529 \times (-8) \approx -4.232 \, \text{Å} \]

2. **For Y (n = 4 to n = 2)**:
\[ \Delta r_Y = r_2 - r_4 = 0.529 \times (2^2 - 4^2) = 0.529 \times (4 - 16) = 0.529 \times (-12) \approx -6.348 \, \text{Å} \]

3. **For Z (n = 5 to n = 3)**:
\[ \Delta r_Z = r_3 - r_5 = 0.529 \times (3^2 - 5^2) = 0.529 \times (9 - 25) = 0.529 \times (-16) \approx -8.464 \, \text{Å} \]

### Conclusion:
- (a) The transition with the shortest wavelength is **X** (n = 3 to n = 1).
- (b) The transition with the longest change in orbit radius is **Z** (n = 5 to n = 3).