To solve the problem step by step, we will follow these calculations:
### Step 1: Calculate the energy of a single photon
The energy of a photon can be calculated using the formula:
\[
E = \frac{1240}{\lambda}
\]
where \(E\) is in electron volts (eV) and \(\lambda\) is in nanometers (nm).
Given:
\(\lambda = 310 \, \text{nm}\)
Substituting the value:
\[
E = \frac{1240}{310} \approx 4 \, \text{eV}
\]
### Step 2: Check if photoemission occurs
The work function \(\phi\) of the metal is given as \(2 \, \text{eV}\). Since the energy of the photon \(4 \, \text{eV}\) is greater than the work function, photoemission will occur.
### Step 3: Calculate the number of photons incident per second
The intensity \(I\) of the light is given as:
\[
I = 6.4 \times 10^{-5} \, \text{W/cm}^2
\]
The power \(P\) incident on the area \(A\) (1 cm²) is:
\[
P = I \times A = 6.4 \times 10^{-5} \times 1 = 6.4 \times 10^{-5} \, \text{W}
\]
The energy of each photon is \(4 \, \text{eV}\), which we need to convert to joules:
\[
1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}
\]
Thus,
\[
E = 4 \, \text{eV} = 4 \times 1.6 \times 10^{-19} = 6.4 \times 10^{-19} \, \text{J}
\]
Now, the number of photons \(n\) incident per second can be calculated using:
\[
n = \frac{P}{E} = \frac{6.4 \times 10^{-5}}{6.4 \times 10^{-19}} = 10^{14} \, \text{photons/second}
\]
### Step 4: Calculate the number of photoelectrons emitted
Given that only \(1\) out of every \(10^3\) photons results in the emission of a photoelectron, the number of photoelectrons emitted per second \(n'\) is:
\[
n' = n \times \text{efficiency} = 10^{14} \times \frac{1}{10^3} = 10^{11}
\]
### Step 5: Find the value of \(x\)
The problem states that the number of photoelectrons emitted in one second is \(10x\). Therefore, we can set up the equation:
\[
10x = 10^{11}
\]
Solving for \(x\):
\[
x = 10^{10}
\]
Thus, the final answer is:
\[
x = 11
\]
