A metal plate of area 1 × 10 - 4   m 2 is illuminated by a radiation of intensity 16   m i l l i   W m 2 . The work function of the metal is 5   e V . The energy of the incident photons is 10   e V and only 10 % of it produces photo electrons. The number of emitted…

Question

A metal plate of area 1 &#215; 10 - 4 &#160; m 2 is illuminated by a radiation of intensity 16 &#160; m i l l i &#160; W m 2 . The work function of the metal is 5 &#160; e V . The energy of the incident photons is 10 &#160; e V and only 10 % of it produces photo electrons. The number of emitted photo electron per second and their maximum energy, respectively, will be: 1 e V = 1.6 &#215; 10 - 19 J

Aicharya · Accepted Answer

Einstein's photoelectric equation
Maximum

K E = \frac{h C}{λ} - ϕ

= 10 - 5

= 5 e V

Intensity

I = \frac{(N_{p}) \frac{h C}{λ}}{t A} \Rightarrow \frac{N_{p}}{t} = \frac{I A}{(\frac{h C}{λ})}

and number of electrons per second.

= (\frac{N_{p}}{t}) \times 10 %

= (\frac{I A}{\frac{h C}{λ}}) \times \frac{1}{10} = \frac{16 \times 10^{- 3} \times 10^{- 4}}{10 \times 1.6 \times 10^{- 19} \times 10}

= 10^{11}

per sec

A metal plate of area 1 × 10 - 4   m 2 is illuminated by a radiation of intensity 16   m i l…

Solution

A metal plate of area 1 &#215; 10 - 4 &#160; m 2 is illuminated by a radiation of intensity 16 &#160; m i l…

Solution

A metal plate of area 1 × 10 - 4 m 2 is illuminated by a radiation of intensity 16 m i l…