A metal plate of area is illuminated by a radiation of intensity . The work function of the metal is . The energy of the incident photons is 10eV and only 1% of it produces photo electrons. The number of emitted photo electrons per second and their maxim

Question

A metal plate of area $2 \times 10^{- 4} m^{2}$ is illuminated by a radiation of intensity $8 m W / m^{2}$ . The work function of the metal is $5 e V$ . The energy of the incident photons is 10eV and only 1% of it produces photo electrons. The number of emitted photo electrons per second and their maximum energy, respectively, will be $[1 e V = 1.6 \times 10^{- 19} J]$

Smart Achievers · Accepted Answer

(4)

$P = I A = 8 \times 10^{- 3} \times 2 \times 10^{- 4} = 16 \times 10^{- 7} W$

$Photons per sec = \frac{P}{h ν} = \frac{16 \times 10^{- 7}}{10 \times 1.6 \times 10^{- 19}} = 10^{12}$

$Photo electrons/sec = 10^{12} \times \frac{1}{100} = 10^{10}$

Solution

1 $10^{14}$ and $10 e V$

2 $10^{12}$ and $5 e V$

3 $10^{11}$ and $5 e V$

4 $10^{10}$ and $5 e V$

Ans.
(4)

$P = I A = 8 \times 10^{- 3} \times 2 \times 10^{- 4} = 16 \times 10^{- 7} W$

$Photons per sec = \frac{P}{h ν} = \frac{16 \times 10^{- 7}}{10 \times 1.6 \times 10^{- 19}} = 10^{12}$

$Photo electrons/sec = 10^{12} \times \frac{1}{100} = 10^{10}$

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Solution

1 1014 and 10 eV

2 1012 and 5 eV

3 1011 and 5 eV

4 1010 and 5 eV