A beam of light of wavelength 400 nm and power 1.55 mW is directed at the cathode of a photoelectric cell. If only 10 % of the incident photons effectively produce photoelectron, then find current due to these electrons. (given, h c=1240 eV - nm , e=1.6 × 10-19 C )

Question

A beam of light of wavelength 400 nm and power 1.55 mW is directed at the cathode of a photoelectric cell. If only 10 % of the incident photons effectively produce photoelectron, then find current due to these electrons. (given, h c=1240 eV - nm , e=1.6 × 10-19 C ) (A) 5 μ A (B) 40 μ A (C) 50 μ A (D) 11.4 μ A

Tardigrade Admin · Accepted Answer

The energy of incident photon of wavelength

400 nm

is

E = \frac{h c}{λ} = \frac{1240}{400} = 3.1 e V

∴

Number of such photons in a beam of light of power

1.55 mW

= \frac{1.55 \times 1 0 ^{- 3}}{3.1 \times 1.6 \times 1 0 ^{- 19}}

= 3.125 \times 1 0^{15}

per second
This implies, number of photons that were used to produce photoelectron per second

= 10%

of

3.125 \times 1 0^{15}

= 3.125 \times 14

per second
Hence, the current due to such photoelectron per second

= 3.125 \times 1 0^{14} \times 1.6 \times 1 0^{- 19}

= 5 \times 1 0^{- 5} A = 50 μ A

Q. A beam of light of wavelength $400 nm$ and power $1.55 mW$ is directed at the cathode of a photoelectric cell. If only $10%$ of the incident photons effectively produce photoelectron, then find current due to these electrons. (given, $h c = 1240 e V - nm, e = 1.6 \times 1 0^{- 19} C$ )

$5 μ A$

$40 μ A$

$50 μ A$

$11.4 μ A$

Q. A beam of light of wavelength 400nm and power 1.55mW is directed at the cathode of a photoelectric cell. If only 10% of the incident photons effectively produce photoelectron, then find current due to these electrons. (given, hc=1240eV−nm,e=1.6×10−19C )

5μA

40μA

50μA

11.4μA

Q. A beam of light of wavelength $400 nm$ and power $1.55 mW$ is directed at the cathode of a photoelectric cell. If only $10%$ of the incident photons effectively produce photoelectron, then find current due to these electrons. (given, $h c = 1240 e V - nm, e = 1.6 \times 1 0^{- 19} C$ )

$5 μ A$

$40 μ A$

$50 μ A$

$11.4 μ A$