Question

a point source of light emits photon of energy 4.8eV at the rate of 10^5​

Answer 1

Answer:

$Energy of emitted photons </p><p>E </p><p>1</p><p> </p><p> =0.5 eV=8×10 </p><p>−19</p><p> </p><p>Energy emitted per second by the source</p><p>E </p><p>2</p><p> </p><p> =3.2×10 </p><p>−3</p><p> J</p><p>No. of photons emitted per second = </p><p>E </p><p>1</p><p> </p><p> </p><p>E </p><p>2</p><p> </p><p> </p><p> </p><p> =4×10 </p><p>15</p><p> </p><p>No. of photons incident on the sphere per second</p><p>n= </p><p>4π(0.8) </p><p>2</p><p> </p><p>4×10 </p><p>15</p><p> </p><p> </p><p> ×π(8×10 </p><p>−3</p><p> ) </p><p>2</p><p> ≈10 </p><p>11</p><p> per second</p><p>No, of photoelectron emitted per second </p><p>n </p><p>′</p><p> = </p><p>10 </p><p>6</p><p> </p><p>10 </p><p>11</p><p> </p><p> </p><p> =10 </p><p>5</p><p> per second</p><p>Max. K.E. of photo electron </p><p>k </p><p>max</p><p> </p><p> = Energy of incident photons work function =2 eV=3.2×10 </p><p>−19</p><p> J</p><p>de-Broglie wavelength of these photoelectrons</p><p>λ </p><p>1</p><p> </p><p> = </p><p>2k </p><p>max</p><p> </p><p> m </p><p>e</p><p> </p><p> </p><p> </p><p> </p><p>h</p><p> </p><p> =8.68 A</p><p>wavelength of incident light </p><p>λ </p><p>2</p><p> </p><p> = </p><p>5</p><p>12375</p><p> </p><p> =2475 A</p><p>The desired ratio is λ </p><p>1</p><p> </p><p> /λ </p><p>2</p><p> </p><p> </p><p>Emission of photoelectrons is stopped when its potential is equal to the stopping potential required of the fastest moving electrons</p><p>k </p><p>max</p><p> </p><p> =2eV</p><p>Stopping potential v </p><p>0</p><p> </p><p> =2 volt</p><p>⇒2=kq/r</p><p>⇒q=1.78×10 </p><p>−12</p><p> C</p><p>⇒t= </p><p>10 </p><p>5</p><p> ×1.6×10 </p><p>−19</p><p> </p><p>q</p><p> </p><p> </p><p>t≈111 seconds.$