Radiation falls on a target kept within a solenoid with 20 turns per cm, carrying a current 2.5 A. Electrons emitted move in a circle with a maximum radius of 1 cm. Find the wavelength of radiation, given that the work function of the target is 0.5 V. `e 1.6xx10^(-19)C`, `h = 6.625xx 10^(-31)J-s`, `m = 9.1 xx 10kg`

Answer

To solve the problem step by step, we will follow the outlined approach based on the information given in the question.

### Step 1: Understand the Problem
We need to find the wavelength of radiation that causes electrons to be emitted from a target placed inside a solenoid. The parameters provided include the work function of the target, the current in the solenoid, the number of turns per unit length, and the maximum radius of the circular motion of the emitted electrons.

### Step 2: Identify the Given Values
- Work function (φ) = 0.5 eV
- Current (I) = 2.5 A
- Number of turns per cm (n) = 20 turns/cm = 2000 turns/m
- Maximum radius of electron motion (r) = 1 cm = 0.01 m
- Charge of electron (e) = \(1.6 \times 10^{-19}\) C
- Planck's constant (h) = \(6.625 \times 10^{-34}\) J·s
- Mass of electron (m) = \(9.1 \times 10^{-31}\) kg

### Step 3: Calculate the Magnetic Field (B)
The magnetic field inside a solenoid is given by the formula:
\[ B = \mu_0 n I \]
Where:
- \(\mu_0 = 4\pi \times 10^{-7} \, \text{T m/A}\)

Substituting the values:
\[ B = (4\pi \times 10^{-7}) \times (2000) \times (2.5) \]
Calculating:
\[ B = (4\pi \times 10^{-7}) \times 5000 \approx 6.2832 \times 10^{-3} \, \text{T} \]

### Step 4: Relate Kinetic Energy and Work Function
The energy of the incident radiation can be expressed as:
\[ E = \phi + KE \]
Where \(KE\) is the kinetic energy of the emitted electrons. The kinetic energy can also be expressed in terms of the radius of the circular motion in the magnetic field:
\[ KE = \frac{mv^2}{2} \]

### Step 5: Find the Velocity (v)
The centripetal force acting on the electron is provided by the magnetic force:
\[ qvB = \frac{mv^2}{r} \]
Rearranging gives:
\[ v = \frac{qBr}{m} \]
Substituting values:
\[ v = \frac{(1.6 \times 10^{-19})(6.2832 \times 10^{-3})(0.01)}{9.1 \times 10^{-31}} \]
Calculating:
\[ v \approx 1.1 \times 10^{11} \, \text{m/s} \]

### Step 6: Calculate Kinetic Energy (KE)
Now substituting \(v\) back into the kinetic energy formula:
\[ KE = \frac{1}{2}mv^2 \]
Calculating:
\[ KE = \frac{1}{2}(9.1 \times 10^{-31})(1.1 \times 10^{11})^2 \]
Calculating:
\[ KE \approx 5.5 \times 10^{-8} \, \text{J} \]

### Step 7: Convert KE to eV
To convert joules to electron volts:
\[ KE \approx \frac{5.5 \times 10^{-8}}{1.6 \times 10^{-19}} \approx 343.75 \, \text{eV} \]

### Step 8: Total Energy
Now, total energy \(E\) can be expressed as:
\[ E = \phi + KE = 0.5 + 343.75 = 344.25 \, \text{eV} \]

### Step 9: Find the Wavelength (λ)
Using the relation \(E = \frac{hc}{\lambda}\):
\[ \lambda = \frac{hc}{E} \]
Substituting the values:
\[ \lambda = \frac{(6.625 \times 10^{-34})(3 \times 10^8)}{344.25 \times 1.6 \times 10^{-19}} \]
Calculating:
\[ \lambda \approx 3.5 \times 10^{-11} \, \text{m} = 35.72 \, \text{Å} \]

### Final Answer
The wavelength of the radiation is approximately **35.72 Å**.