Radiations of two photon’s energy, twice and ten times the work function of metal are incident on the metal surface successsively. The ratio of maximum velocities of photoelectrons emitted in two cases is

Options

A. `1:2`
B. `1:3`
C. `1:4`
D. `1:1`

Answer

To solve the problem of finding the ratio of maximum velocities of photoelectrons emitted when two photons with energies twice and ten times the work function of a metal are incident on the metal surface, we can follow these steps:

### Step 1: Understand the Photoelectric Effect
The photoelectric effect states that when light (photons) hits a metal surface, it can eject electrons if the energy of the photons is greater than the work function (φ) of the metal. The maximum kinetic energy (K.E.) of the emitted photoelectrons can be given by the equation:
\[ K.E. = E - \phi \]
where \( E \) is the energy of the incident photon.

### Step 2: Define Energies for Each Case
1. For the first case, the energy of the photon is twice the work function:
\[ E_1 = 2\phi \]
Therefore, the maximum kinetic energy of the emitted photoelectrons is:
\[ K.E_1 = E_1 - \phi = 2\phi - \phi = \phi \]

2. For the second case, the energy of the photon is ten times the work function:
\[ E_2 = 10\phi \]
Thus, the maximum kinetic energy of the emitted photoelectrons is:
\[ K.E_2 = E_2 - \phi = 10\phi - \phi = 9\phi \]

### Step 3: Relate Kinetic Energy to Velocity
The maximum kinetic energy can also be expressed in terms of the maximum velocity (v_max) of the emitted electrons:
\[ K.E = \frac{1}{2} m v_{max}^2 \]
where \( m \) is the mass of the electron.

### Step 4: Set Up Equations for Maximum Velocities
From the kinetic energy expressions, we can write:
1. For the first case:
\[ K.E_1 = \frac{1}{2} m v_{max1}^2 = \phi \]
Rearranging gives:
\[ v_{max1}^2 = \frac{2\phi}{m} \]

2. For the second case:
\[ K.E_2 = \frac{1}{2} m v_{max2}^2 = 9\phi \]
Rearranging gives:
\[ v_{max2}^2 = \frac{18\phi}{m} \]

### Step 5: Find the Ratio of Maximum Velocities
Now we can find the ratio of the maximum velocities:
\[ \frac{v_{max1}}{v_{max2}} = \sqrt{\frac{v_{max1}^2}{v_{max2}^2}} = \sqrt{\frac{\frac{2\phi}{m}}{\frac{18\phi}{m}}} = \sqrt{\frac{2}{18}} = \sqrt{\frac{1}{9}} = \frac{1}{3} \]

### Conclusion
Thus, the ratio of the maximum velocities of photoelectrons emitted in the two cases is:
\[ \frac{v_{max1}}{v_{max2}} = \frac{1}{3} \]