The velocity of the most energetic electrons emitted from a metallic surface is doubled when the frequecy v of incident radiation is double. The work function of this metal is

Options

A. Zero
B. `hv//3`
C. `hv//2`
D. `2hv//3`

Explanation

To solve the problem, we will use the principles of the photoelectric effect and Einstein's photoelectric equation. Let's break down the solution step by step.

### Step 1: Understand the Photoelectric Effect
According to Einstein's photoelectric equation, the maximum kinetic energy (K.E.) of the emitted electrons is given by:
\[ K.E. = h\nu - \phi_0 \]
where:
- \( K.E. \) is the maximum kinetic energy of the emitted electrons,
- \( h \) is Planck's constant,
- \( \nu \) is the frequency of the incident radiation,
- \( \phi_0 \) is the work function of the metal.

### Step 2: Write the Initial Condition
Let the initial frequency of the incident radiation be \( \nu \) and the maximum velocity of the emitted electrons be \( V_{max} \). The maximum kinetic energy can be expressed as:
\[ K.E. = \frac{1}{2} m V_{max}^2 = h\nu - \phi_0 \]
This is our **Equation 1**.

### Step 3: Write the Condition After Doubling the Frequency
When the frequency is doubled, the new frequency becomes \( 2\nu \) and the new maximum velocity becomes \( 2V_{max} \). The maximum kinetic energy can now be expressed as:
\[ K.E. = \frac{1}{2} m (2V_{max})^2 = h(2\nu) - \phi_0 \]
Simplifying this gives:
\[ K.E. = \frac{1}{2} m (4V_{max}^2) = 4 \left(\frac{1}{2} m V_{max}^2\right) \]
Thus, we can write:
\[ 2m V_{max}^2 = 2h\nu - \phi_0 \]
This is our **Equation 2**.

### Step 4: Relate the Two Equations
From **Equation 1**, we have:
\[ \frac{1}{2} m V_{max}^2 = h\nu - \phi_0 \]
From **Equation 2**, we can express it as:
\[ 2 \left(h\nu - \phi_0\right) = 2h\nu - \phi_0 \]
Substituting \( \frac{1}{2} m V_{max}^2 \) from **Equation 1** into **Equation 2** gives:
\[ 4(h\nu - \phi_0) = 2h\nu - \phi_0 \]

### Step 5: Simplify the Equation
Expanding and simplifying:
\[ 4h\nu - 4\phi_0 = 2h\nu - \phi_0 \]
Rearranging terms:
\[ 4h\nu - 2h\nu = 4\phi_0 - \phi_0 \]
This simplifies to:
\[ 2h\nu = 3\phi_0 \]

### Step 6: Solve for the Work Function
Now, we can solve for the work function \( \phi_0 \):
\[ \phi_0 = \frac{2}{3} h\nu \]

### Conclusion
Thus, the work function of the metal is:
\[ \phi_0 = \frac{2}{3} h\nu \]