The collector of the photocell (in photoelectric experiment) is made of tungsten while the emitter is of platinum having work function of 10 eV. Monochromatic radiation of wavelength 124 mathringA and power 100 watt is incident on emitter which emits photo electrons with a quantum efficiency of 1 % . The accelerating voltage across the photocell is of 10,000 volts (Use: h c =12400 eV mathringA) If the source of monochromatic radiation of wavelength 124 mathringAhas an efficiency of 50 %, and the power of X-ray emitted by the tungsten target is 3 W, the overall efficiency of the apparatus for X-ray production is

Question

The collector of the photocell (in photoelectric experiment) is made of tungsten while the emitter is of platinum having work function of 10 eV. Monochromatic radiation of wavelength 124 mathringA and power 100 watt is incident on emitter which emits photo electrons with a quantum efficiency of 1 % . The accelerating voltage across the photocell is of 10,000 volts (Use: h c =12400 eV mathringA) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/6c9c360ee4888473414a24a40ee01937-.png" /> If the source of monochromatic radiation of wavelength 124 mathringAhas an efficiency of 50 %, and the power of X-ray emitted by the tungsten target is 3 W, the overall efficiency of the apparatus for X-ray production is (A) 1 % (B) 0.1 % (C) 1.5 % (D) 0.67 %

Tardigrade Admin · Accepted Answer

Power provided by accelerating potential

= 100 W

Power consumed by the source of

124 \overset{˚}{A} = \frac{100}{( 0.5 )} = 200 W

Overall efficiency

= \frac{net power of X - ray}{net power supplied}

= \frac{3 W}{100 W + 200 W}

= 0.01 = 1%

$1%$

$0.1%$

$1.5%$

$0.67%$

Power provided by accelerating potential $= 100 W$
Power consumed by the source of
$124 \overset{˚}{A} = \frac{100}{( 0.5 )} = 200 W$
Overall efficiency
$= \frac{net power of X - ray}{net power supplied}$
$= \frac{3 W}{100 W + 200 W}$
$= 0.01 = 1%$

1%

0.1%

1.5%

0.67%

Power provided by accelerating potential =100W Power consumed by the source of 124A˚=(0.5)100​=200W Overall efficiency = net power supplied net power of X− ray ​ =100W+200W3W​ =0.01=1%

$1%$

$0.1%$

$1.5%$

$0.67%$

Power provided by accelerating potential $= 100 W$
Power consumed by the source of
$124 \overset{˚}{A} = \frac{100}{( 0.5 )} = 200 W$
Overall efficiency
$= \frac{net power of X - ray}{net power supplied}$
$= \frac{3 W}{100 W + 200 W}$
$= 0.01 = 1%$