A photon with energy of 4.9 eV ejects photoelectrons from tungsten. When the ejcect phototelectrons from tungsten. When the ejected electron enters a cnstant magnetic field of strangth B=2.5 mT at an angle of `60^@` with the field direction, the maximum pitch fo the helix described by the electron is found to be 2.7 mm. Find the work function of the metal in electron volt. Given that specific charge of electron is `1.76xx10^11 C/kg.

Explanation

To solve the problem, we need to find the work function of tungsten given the energy of the photon, the magnetic field strength, the angle, and the maximum pitch of the helix described by the electron.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Energy of the photon, \( E = 4.9 \, \text{eV} \)
- Magnetic field strength, \( B = 2.5 \, \text{mT} = 2.5 \times 10^{-3} \, \text{T} \)
- Angle with the magnetic field, \( \theta = 60^\circ \)
- Maximum pitch of the helix, \( P = 2.7 \, \text{mm} = 2.7 \times 10^{-3} \, \text{m} \)
- Specific charge of the electron, \( \alpha = 1.76 \times 10^{11} \, \text{C/kg} \)

2. **Formula for Pitch:**
The pitch \( P \) of the helix can be expressed as:
\[ P = V \cos \theta \cdot t \]
where \( V \) is the velocity of the electron and \( t \) is the time period of the motion.

3. **Time Period of Motion:**
The time period \( t \) can be calculated using:
\[ t = \frac{2\pi m}{QB} \]
where \( Q \) is the charge of the electron and \( m \) is the mass of the electron. The specific charge \( \alpha \) can be expressed as \( \alpha = \frac{Q}{m} \), thus:
\[ t = \frac{2\pi}{B \alpha} \]

4. **Substituting for \( P \):**
Substitute \( t \) into the pitch formula:
\[ P = V \cos \theta \cdot \frac{2\pi}{B \alpha} \]
Rearranging gives:
\[ V = \frac{P B \alpha}{2\pi \cos \theta} \]

5. **Calculating Velocity \( V \):**
Substitute the known values:
\[ V = \frac{(2.7 \times 10^{-3}) \cdot (2.5 \times 10^{-3}) \cdot (1.76 \times 10^{11})}{2\pi \cdot \cos(60^\circ)} \]
Since \( \cos(60^\circ) = \frac{1}{2} \):
\[ V = \frac{(2.7 \times 10^{-3}) \cdot (2.5 \times 10^{-3}) \cdot (1.76 \times 10^{11})}{\pi} \]

6. **Calculating Kinetic Energy:**
The maximum kinetic energy \( KE \) of the ejected electron is given by:
\[ KE = \frac{1}{2} m V^2 \]
The work function \( W \) can be found using:
\[ W = E - KE \]
where \( E \) is the energy of the photon.

7. **Substituting for \( KE \):**
Substitute \( V \) into the kinetic energy formula:
\[ KE = \frac{1}{2} m \left(\frac{P B \alpha}{2\pi \cos \theta}\right)^2 \]

8. **Finding Work Function \( W \):**
Finally, substitute the expression for \( KE \) back into the equation for \( W \):
\[ W = E - \frac{1}{2} m \left(\frac{P B \alpha}{2\pi \cos \theta}\right)^2 \]

9. **Calculating Work Function:**
Substitute the known values into the equation to find \( W \).

### Final Calculation:
After substituting all values and calculating, we find:
\[ W \approx 4.5 \, \text{eV} \]