Ultraviolet light of wavelength 300nm and intensity `1.0Wm^-2` falls on the surface of a photosensitive material. If one per cent of the incident photons produce photoelectrons, then the number of photoelectrons emitted per second from an area of 1.0 `cm^2` of the surface is nearly

Options

A. `9.61xx10^14s^-1`
B. `4.12xx10^13s^-1`
C. `1.51xx10^12s^-1`
D. `2.13xx10^11s^-1`

Explanation

To solve the problem step by step, we will follow the procedure outlined in the video transcript.

### Step 1: Convert Wavelength to Meters
The wavelength of the ultraviolet light is given as 300 nm. We need to convert this to meters.

\[ \text{Wavelength} (\lambda) = 300 \, \text{nm} = 300 \times 10^{-9} \, \text{m} \]

**Hint:** Remember that 1 nm = \(10^{-9}\) m.

### Step 2: Use the Intensity Formula to Find Number of Photons
The intensity \(I\) of the light is given as \(1.0 \, \text{W/m}^2\). The formula relating intensity, number of photons \(n\), Planck's constant \(h\), speed of light \(c\), and wavelength \(\lambda\) is:

\[ I = n \cdot \frac{hc}{\lambda} \]

Rearranging this formula to solve for \(n\):

\[ n = \frac{I \cdot \lambda}{h \cdot c} \]

Substituting the known values:
- \(I = 1.0 \, \text{W/m}^2\)
- \(\lambda = 300 \times 10^{-9} \, \text{m}\)
- \(h = 6.63 \times 10^{-34} \, \text{Js}\)
- \(c = 3.0 \times 10^{8} \, \text{m/s}\)

\[ n = \frac{(1.0) \cdot (300 \times 10^{-9})}{(6.63 \times 10^{-34}) \cdot (3.0 \times 10^{8})} \]

Calculating this gives:

\[ n \approx 1.51 \times 10^{18} \, \text{photons/m}^2/\text{s} \]

**Hint:** Make sure to keep track of units when substituting values.

### Step 3: Calculate the Number of Photoelectrons
According to the problem, only 1% of the incident photons produce photoelectrons. Therefore, the number of photoelectrons emitted per second \(n'\) is:

\[ n' = 0.01 \cdot n \]

Substituting the value of \(n\):

\[ n' = 0.01 \cdot (1.51 \times 10^{18}) = 1.51 \times 10^{16} \, \text{photoelectrons/m}^2/\text{s} \]

**Hint:** To find 1% of a number, multiply by 0.01.

### Step 4: Convert to cm²
We need to convert the number of photoelectrons from per square meter to per square centimeter. Since \(1 \, \text{m}^2 = 10^4 \, \text{cm}^2\):

\[ n' = \frac{1.51 \times 10^{16}}{10^4} = 1.51 \times 10^{12} \, \text{photoelectrons/cm}^2/\text{s} \]

**Hint:** Remember that there are \(10^4\) cm² in 1 m².

### Final Answer
The number of photoelectrons emitted per second from an area of \(1.0 \, \text{cm}^2\) of the surface is approximately:

\[ 1.51 \times 10^{12} \, \text{photoelectrons/s} \]