`14xx10^(15)` electrons reach the anode per second. If the power consumed is 448 milliwatts, then the plate (anode) voltage is

14 xx 10^15 electrons reach the anode per second. If the power consumed is 448 milliwatts, then the anode voltage will be :

An electron is revolving n times per second. The charge passing in t second is

The cathode its 8 xx 10^(14) electrons per second, when heated. When 400 V is applied to anode all the emitted electrons reach the anode. The charge on electron is 1.6 xx 10^(-19)C . The maximum anode current is

A cathode emits 1.8xx10^(17) electron per second and all the electrons reach the anode when it Is given a positive potential of 400V . Given e=1.6xx10^(-19) C, the maximum anode current is .

The maximum energy of the photoelectrons emitted in a photocell Is 5eV. For no photoelectron to reach the anode, the potential difference of the anode with respect to the photo sensitive plate should be :