Assuming photoemission to take place , the factor by which the maximum velocity of the emitted photoelectrons changes when the wavelength of the incident radiation is increased four times , is

Options

A. 4
B. `1/4`
C. `2`
D. `1/2`

Answer

To solve the problem step by step, we will analyze the relationship between the wavelength of the incident radiation and the maximum velocity of the emitted photoelectrons.

### Step 1: Understand the relationship between kinetic energy and wavelength
The maximum kinetic energy (K.E.) of the emitted photoelectrons can be expressed using the equation:
\[ K.E. = \frac{1}{2} m V_m^2 = \frac{hc}{\lambda} - \phi \]
where:
- \( m \) is the mass of the electron,
- \( V_m \) is the maximum velocity of the emitted photoelectrons,
- \( h \) is Planck's constant,
- \( c \) is the speed of light,
- \( \lambda \) is the wavelength of the incident radiation,
- \( \phi \) is the work function of the material.

### Step 2: Neglect the work function
Assuming that the energy of the incident photons is much greater than the work function (i.e., \( \frac{hc}{\lambda} \gg \phi \)), we can neglect \( \phi \):
\[ \frac{1}{2} m V_m^2 = \frac{hc}{\lambda} \]

### Step 3: Rearranging the equation
From the above equation, we can express \( V_m \):
\[ V_m^2 = \frac{2hc}{m\lambda} \]
This shows that \( V_m^2 \) is inversely proportional to \( \lambda \):
\[ V_m^2 \propto \frac{1}{\lambda} \]

### Step 4: Relate the velocities for different wavelengths
Let \( V_{m1} \) be the maximum velocity for the initial wavelength \( \lambda_1 \) and \( V_{m2} \) for the new wavelength \( \lambda_2 \). We can write:
\[ \frac{V_{m1}^2}{V_{m2}^2} = \frac{\lambda_2}{\lambda_1} \]

### Step 5: Substitute the new wavelength
Given that the wavelength is increased four times, we have:
\[ \lambda_2 = 4\lambda_1 \]
Substituting this into the equation gives:
\[ \frac{V_{m1}^2}{V_{m2}^2} = \frac{4\lambda_1}{\lambda_1} = 4 \]

### Step 6: Solve for the ratio of velocities
Taking the square root of both sides, we find:
\[ \frac{V_{m1}}{V_{m2}} = \sqrt{4} = 2 \]
This implies:
\[ V_{m2} = \frac{V_{m1}}{2} \]

### Step 7: Determine the factor of change
The factor by which the maximum velocity changes is:
\[ \text{Factor} = \frac{V_{m2}}{V_{m1}} = \frac{V_{m1}/2}{V_{m1}} = \frac{1}{2} \]

### Conclusion
Thus, the factor by which the maximum velocity of the emitted photoelectrons changes when the wavelength of the incident radiation is increased four times is:
\[ \frac{1}{2} \]

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