Threshold frequency for a certain metal is `v_0`. When light of frequency `2v_0` is incident on it, the maximum velocity of photoelectrons is `4xx10^8 cm s^-1`. If frequency of incident radiation is increased to `5v_0`, then the maximum velocity of photoelectrons, in cm `s^-1`, will be

Options

A. `(4)/(5)xx10^8`
B. `2xx10^8`
C. `8xx10^8`
D. `20xx10^8`

Explanation

To solve the problem, we will use Einstein's photoelectric equation, which relates the energy of the incident photons to the kinetic energy of the emitted photoelectrons. The equation is given by:

\[ K.E. = h\nu - h\nu_0 \]

Where:
- \( K.E. \) is the maximum kinetic energy of the photoelectrons,
- \( h \) is Planck's constant,
- \( \nu \) is the frequency of the incident light,
- \( \nu_0 \) is the threshold frequency.

The kinetic energy can also be expressed in terms of the maximum velocity \( v \) of the photoelectrons:

\[ K.E. = \frac{1}{2} mv^2 \]

Where:
- \( m \) is the mass of the photoelectron,
- \( v \) is the maximum velocity of the photoelectrons.

### Step 1: Calculate the maximum kinetic energy when frequency is \( 2\nu_0 \)

Given that the frequency of the incident light is \( 2\nu_0 \) and the maximum velocity of the photoelectrons is \( 4 \times 10^8 \, \text{cm/s} \):

Using the photoelectric equation:

\[ K.E. = h(2\nu_0) - h\nu_0 = h\nu_0 \]

Now, substituting for kinetic energy:

\[ h\nu_0 = \frac{1}{2} m (4 \times 10^8)^2 \]

### Step 2: Simplify the equation

Calculating the right side:

\[ h\nu_0 = \frac{1}{2} m \cdot 16 \times 10^{16} = 8m \times 10^{16} \]

Thus, we have:

\[ h\nu_0 = 8m \times 10^{16} \quad \text{(Equation 1)} \]

### Step 3: Calculate the maximum kinetic energy when frequency is \( 5\nu_0 \)

Now, when the frequency of the incident radiation is increased to \( 5\nu_0 \):

Using the photoelectric equation again:

\[ K.E. = h(5\nu_0) - h\nu_0 = 4h\nu_0 \]

Now substituting for kinetic energy:

\[ 4h\nu_0 = \frac{1}{2} mv'^2 \]

### Step 4: Substitute \( h\nu_0 \) from Equation 1

From Equation 1, we know:

\[ h\nu_0 = 8m \times 10^{16} \]

Substituting this into the equation:

\[ 4(8m \times 10^{16}) = \frac{1}{2} mv'^2 \]

This simplifies to:

\[ 32m \times 10^{16} = \frac{1}{2} mv'^2 \]

### Step 5: Cancel \( m \) and solve for \( v' \)

Canceling \( m \) from both sides (assuming \( m \neq 0 \)):

\[ 32 \times 10^{16} = \frac{1}{2} v'^2 \]

Multiplying both sides by 2:

\[ 64 \times 10^{16} = v'^2 \]

Taking the square root:

\[ v' = \sqrt{64 \times 10^{16}} = 8 \times 10^8 \, \text{cm/s} \]

### Final Answer

The maximum velocity of photoelectrons when the frequency of incident radiation is increased to \( 5\nu_0 \) is:

\[ \boxed{8 \times 10^8 \, \text{cm/s}} \]