Answer to Question #90852 in Quantum Mechanics for allu arjun

Question

A photocell in the saturation mode is irradiated
by light of wavelength  = 6600 Å. The
corresponding spectral sensitivity of the cell is
s = 4.8 mA/W. Find the yield of
photoelectrons, i.e. the number of
photoelectrons produced by each incident
photon. [Take : h = 6.6 × 10–34 J-s]

Answer 1

The energy of a single photon can be calculated as

$E=\frac{h c}{\lambda}$

The spectral sensitivity can be defined in terms of either current and power or charge and energy as:

$s_{\lambda} = \frac{I}{P} = \frac{I \cdot t}{P \cdot t} = \frac{q}{E} = \frac{N e}{E}$

Hence, the number of photoelectrons produced by each incident photon is

$N = \frac{s_\lambda E}{e} = \frac{s_\lambda h c}{e \lambda}$

Substituting the numerical values, we obtain:

$N = \frac{4.8 \cdot 10^{-3} \cdot 6.6 \cdot 10^{-34} \cdot 3 \cdot 10^8}{6.6 \cdot 10^3 \cdot 10^{-10} \cdot 1.6 \cdot 10^{-19}} \approx 9 \cdot 10^{-3} = 0.009$

Answer: 0.009