A source emits monochromatic light of frequency  $5.5\times {10}^{14}Hz$ at a rate of $0.1W$. Of the photons given out, $0.15\%$ fall on the cathode of a photocell which gives a current of $6\mu A$ in an external circuit.

 Find the energy of a photon.

To find the energy of a photon, we use the formula:

E = h × f

Where:

E = Energy of the photon

h = Planck's constant = 6.63 × 10^-34 J·s

f = Frequency of the light = 5.5 × 10¹⁴ Hz

Step 1: Calculate Energy

Substitute the values:

E = 6.63 × 10^-34 × 5.5 × 10¹⁴

E = 3.6465 × 10^-19 J

Step 2: Convert Energy to Electron Volts (eV)

1 eV = 1.6 × 10^-19 J, so:

E_eV = 3.6465 × 10^-19 ÷ 1.6 × 10^-19

E_eV = 2.27 eV