A metal plate of area 1 × 10-4 m2 is illuminated by a radiation of intensity 16 mW/m2. The work function of the metal is 5 eV. The energy of the incident photons is 10 eV and only 10 % of it produces photo electrons. The number of emitted photo electrons per second and their maximum energy, respectively, will be: [1 eV = 1.6 × 10-19 J]

Question

A metal plate of area 1 × 10-4 m2 is illuminated by a radiation of intensity 16 mW/m2. The work function of the metal is 5 eV. The energy of the incident photons is 10 eV and only 10 % of it produces photo electrons. The number of emitted photo electrons per second and their maximum energy, respectively, will be: [1 eV = 1.6 × 10-19 J] (A) 1010 and 5 ev  (B) 1014 and 10 ev  (C) 1012 and 5 ev  (D) 1011 and 5 ev

Tardigrade Admin · Accepted Answer

I =(nE/At) 16×10-3 =((n/t)) textPhoton (10×1.6 ×10-19/10-4) =1012

$1 0^{10}$ and $5 e v$

$1 0^{14}$ and $10 e v$

$1 0^{12}$ and $5 e v$

$1 0^{11}$ and $5 e v$

$I = \frac{n E}{A t}$
$16 \times 1 0^{- 3} = (\frac{n}{t})_{Photon} \frac{10 \times 1.6 \times 1 0 ^{- 19}}{1 0 ^{- 4}} = 1 0^{12}$

1010 and 5ev

1014 and 10ev

1012 and 5ev

1011 and 5ev

I=AtnE​ 16×10−3=(tn​)Photon​10−410×1.6×10−19​=1012

$1 0^{10}$ and $5 e v$

$1 0^{14}$ and $10 e v$

$1 0^{12}$ and $5 e v$

$1 0^{11}$ and $5 e v$

$I = \frac{n E}{A t}$
$16 \times 1 0^{- 3} = (\frac{n}{t})_{Photon} \frac{10 \times 1.6 \times 1 0 ^{- 19}}{1 0 ^{- 4}} = 1 0^{12}$