The eye can detect `5xx10^(4)` photons `(m^2s)^(-1)` of green light (`lamda=5000A)`, whole ear can detect `10^(-13)Wm^(2)` . As a power detector, which is more sensitive and by what factor?

Options

A. Eye is more sensitive and by a factor of 5.00
B. Ear is more sensitive by a factor of 5.00
C. Both are equally sensitive
D. Eye is more sensitive by a factor of `10^(-1)`

Explanation

To determine which is more sensitive between the eye and the ear, we need to calculate the intensity of light detected by the eye and compare it with the intensity detected by the ear.

### Step-by-Step Solution:

1. **Given Data:**
- Photons detected by the eye: \( N = 5 \times 10^4 \) photons \( (m^2 s)^{-1} \)
- Wavelength of green light: \( \lambda = 5000 \, \text{Å} = 5000 \times 10^{-10} \, \text{m} \)
- Intensity detected by the ear: \( I_{\text{ear}} = 10^{-13} \, W/m^2 \)

2. **Calculate Energy of a Single Photon:**
The energy \( E \) of a single photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
Where:
- \( h = 6.63 \times 10^{-34} \, J \cdot s \) (Planck's constant)
- \( c = 3 \times 10^8 \, m/s \) (speed of light)
- \( \lambda = 5000 \times 10^{-10} \, m \)

Substituting the values:
\[ E = \frac{(6.63 \times 10^{-34})(3 \times 10^8)}{5000 \times 10^{-10}} \]
\[ E = \frac{1.989 \times 10^{-25}}{5 \times 10^{-7}} = 3.978 \times 10^{-19} \, J \]

3. **Calculate Intensity Detected by the Eye:**
The intensity \( I_{\text{eye}} \) can be calculated using the formula:
\[ I_{\text{eye}} = N \times E \]
Where \( N = 5 \times 10^4 \) photons \( (m^2 s)^{-1} \).

Substituting the values:
\[ I_{\text{eye}} = (5 \times 10^4) \times (3.978 \times 10^{-19}) \]
\[ I_{\text{eye}} = 1.989 \times 10^{-14} \, W/m^2 \]

4. **Compare the Sensitivity:**
Now we compare the two intensities:
- Intensity detected by the ear: \( I_{\text{ear}} = 10^{-13} \, W/m^2 \)
- Intensity detected by the eye: \( I_{\text{eye}} = 1.989 \times 10^{-14} \, W/m^2 \)

To find out which is more sensitive, we calculate the factor:
\[ \text{Sensitivity Factor} = \frac{I_{\text{ear}}}{I_{\text{eye}}} \]
\[ \text{Sensitivity Factor} = \frac{10^{-13}}{1.989 \times 10^{-14}} \approx 5.03 \]

### Conclusion:
The ear is more sensitive than the eye by a factor of approximately 5.