Question

0 12/30 Eye can detect a signal (colour) of 900 nm, if
total energy of such photons exceeds 5.5 x 10-13 J.
How many minimum photons has to strike on eye to
detect signal? [Take, h = 6.6x 10-34 Js]​

Answer 1

Given:

The wavelength of photon, λ = 900 n m = 9 × 10⁻⁷ m

The total energy of potons, T.E= 5.5 x 10⁻¹³ J

h = 6.6 x 10⁻³⁴ Js

To Find:

The minimum no of photons required to strike on eye to  detect the signal.

Calculation:

- The formula for energy of one photon is given as:

E = hc / λ

⇒ E = 6.6 x 10⁻³⁴ x 3 x 10⁸ / 9 × 10⁻⁷

⇒ E = 2.2 x 10⁻¹⁹ J

- Minimum no of photons = T.E / E

⇒ n = 5.5 x 10⁻¹³ / 2.2 x 10⁻¹⁹

⇒ n = 2.5 x 10⁶

- So, minimum 2.5 x 10⁶ photons are required to strike on eye to  detect the signal.

Answer 2

Given:

Wavelength,

W = 900 nm

   = 9 × 10⁻⁷ m

Photons total energy,

T.E = 5.5 x 10⁻¹³ J

h = 6.6 x 10⁻³⁴ Js

To Find:

Number of photons = ?

Solution:

As we know,

⇒  $E=\frac{hc}{\lambda}$

On putting the estimated values, we get

⇒      $=\frac{6.6\times 10^{-34}\times 3\times 10^8}{9\times 10^{-7}}$

⇒      $=2.2\times 10^{19} \ J$

Now,

⇒  $Number \ of \ photons=\frac{T.E}{E}$

⇒                                   $=\frac{5.5\times 10^{-13}}{2.2\times 10^{-19}}$

⇒                                   $=2.5\times 10^6$

So that the minimum number of photons will be "2.5 x 10⁶".