A photon of energy E ejects a photoelectron from a metel surface whose work function is `phi_(0)`. If this electron enters into a unifrom magnetic field of induction B in a direction perpendicular to the field and describes a circular path of radius r, then the radius r, is given by, (in the usual notation)

Options

A. `sqrt((2m(E-phi_(0)))/(eB))`
B. `sqrt((2m(E-phi_(0)))eB)`
C. `sqrt((2m(E-phi_(0)))/(mB))`
D. `sqrt(2m(E-phi_(0)))/(eB)`

Explanation

To solve the problem step by step, we will use the concepts of photoelectric effect and motion of charged particles in a magnetic field.

### Step 1: Understanding the Photoelectric Effect
When a photon of energy \( E \) strikes a metal surface, it can eject an electron if the energy of the photon is greater than the work function \( \phi_0 \) of the metal. According to Einstein's photoelectric equation:
\[ E = \phi_0 + K_{\text{max}} \]
where \( K_{\text{max}} \) is the maximum kinetic energy of the emitted electron.

### Step 2: Expressing Maximum Kinetic Energy
From the equation above, we can express the maximum kinetic energy of the emitted electron as:
\[ K_{\text{max}} = E - \phi_0 \]

### Step 3: Motion of the Electron in a Magnetic Field
Once the electron is emitted, it enters a uniform magnetic field \( B \) perpendicular to its velocity. The force acting on the electron due to the magnetic field causes it to move in a circular path. The radius \( r \) of this circular path is given by the formula:
\[ r = \frac{mv}{qB} \]
where:
- \( m \) is the mass of the electron,
- \( v \) is the velocity of the electron,
- \( q \) is the charge of the electron (which is \( e \)),
- \( B \) is the magnetic field strength.

### Step 4: Relating Velocity to Kinetic Energy
The kinetic energy of the electron is also given by:
\[ K = \frac{1}{2} mv^2 \]
From this, we can express the velocity \( v \) in terms of kinetic energy:
\[ v = \sqrt{\frac{2K}{m}} \]
Substituting \( K_{\text{max}} \) for \( K \):
\[ v = \sqrt{\frac{2(E - \phi_0)}{m}} \]

### Step 5: Substituting Velocity into the Radius Formula
Now we substitute \( v \) back into the radius formula:
\[ r = \frac{m \cdot \sqrt{\frac{2(E - \phi_0)}{m}}}{eB} \]
This simplifies to:
\[ r = \frac{m \sqrt{2(E - \phi_0)}}{eB \sqrt{m}} = \frac{\sqrt{2m(E - \phi_0)}}{eB} \]

### Final Expression for the Radius
Thus, the radius \( r \) of the circular path of the photoelectron in the magnetic field is given by:
\[ r = \frac{\sqrt{2m(E - \phi_0)}}{eB} \]