A photosensitive metallic surface has work function , `h v_(0)`. If photons of energy ` 2hv_(0)` fall on this surface , the electrons come out with a maximum velocity of `4 xx 10^(6) m//s`. When the photon energy is increased to `5hv_(0)` ,then the maximum velocity of photoelectrons will be

Options

A. `2 xx 10^(6) m//s`
B. `2 xx 10^(7)m//s`
C. `8 xx 10^(5) m//s`
D. `8 xx 10^(6) m//s`

Answer

To solve the problem, we will use the principles of the photoelectric effect and the kinetic energy of the emitted electrons. Let's break down the solution step by step.

### Given:
- Work function of the metallic surface: \( \phi = h \nu_0 \)
- Energy of the first photon: \( E_1 = 2h \nu_0 \)
- Maximum velocity of electrons when \( E_1 \) is used: \( v_1 = 4 \times 10^6 \, \text{m/s} \)
- Energy of the second photon: \( E_2 = 5h \nu_0 \)

### Step 1: Write the kinetic energy equation for the first case
The kinetic energy (KE) of the emitted electrons can be expressed as:
\[ KE = E - \phi \]
For the first case, substituting the values:
\[ KE_1 = E_1 - \phi = 2h \nu_0 - h \nu_0 = h \nu_0 \]
The kinetic energy can also be expressed in terms of velocity:
\[ KE_1 = \frac{1}{2} mv_1^2 \]
Thus, we have:
\[ h \nu_0 = \frac{1}{2} m (4 \times 10^6)^2 \]

### Step 2: Calculate the kinetic energy for the first case
Calculating \( v_1^2 \):
\[ v_1^2 = (4 \times 10^6)^2 = 16 \times 10^{12} \, \text{m}^2/\text{s}^2 \]
Substituting this back into the kinetic energy equation:
\[ h \nu_0 = \frac{1}{2} m (16 \times 10^{12}) \]
This simplifies to:
\[ h \nu_0 = 8 m \times 10^{12} \quad \text{(Equation 1)} \]

### Step 3: Write the kinetic energy equation for the second case
For the second case, the energy of the photon is \( E_2 = 5h \nu_0 \):
\[ KE_2 = E_2 - \phi = 5h \nu_0 - h \nu_0 = 4h \nu_0 \]
Thus, we can express this as:
\[ KE_2 = \frac{1}{2} mv_2^2 \]
This gives us:
\[ 4h \nu_0 = \frac{1}{2} mv_2^2 \quad \text{(Equation 2)} \]

### Step 4: Relate Equation 1 and Equation 2
From Equation 1, we have:
\[ h \nu_0 = 8 m \times 10^{12} \]
Substituting \( h \nu_0 \) from Equation 1 into Equation 2:
\[ 4(8 m \times 10^{12}) = \frac{1}{2} mv_2^2 \]
This simplifies to:
\[ 32 m \times 10^{12} = \frac{1}{2} mv_2^2 \]

### Step 5: Solve for \( v_2^2 \)
Cancelling \( m \) from both sides (assuming \( m \neq 0 \)):
\[ 32 \times 10^{12} = \frac{1}{2} v_2^2 \]
Multiplying both sides by 2:
\[ 64 \times 10^{12} = v_2^2 \]

### Step 6: Calculate \( v_2 \)
Taking the square root:
\[ v_2 = \sqrt{64 \times 10^{12}} = 8 \times 10^6 \, \text{m/s} \]

### Final Answer:
The maximum velocity of the photoelectrons when the photon energy is increased to \( 5h \nu_0 \) is:
\[ v_2 = 8 \times 10^6 \, \text{m/s} \]

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