Discussion Forum

In a photoelectric experiment , a parallel beam of monochromatic light with power of 200 W is incident on a perfectly absorbing cathode of work function 6.25 eV  . The frequency of light is just above the threshold frequency, so that the photoelectrons are emitted with negligible kinetic energy.  Assume that the photoelectron emission  efficiency is 100% . A potential difference of 500 V  is applied between the cathode and the anode. All the emitted electrons are incident normally on the anode and are absorbed. The anode experiences a force F= n × 10^-4 N due to the impact of the eletrons^.  The value of n is .................(Take mass of the electron, m_e =9× 10^-31 kg and eV = 1.6 × 10^-19 J)

Answer:

Explanation:

$\therefore$  Power= nhf

       (where , n=number of photons incident per second)

Since, KE= 0, hf = work-function W

    200=nW= n[6.25×1.6 × 10^-19 ]

$\Rightarrow$    $n=\frac{200}{1.6\times10^{-19}\times6.25}$

  As photon is just above threshold frequency KE_max is zero and they are accelerated by potential difference of 500V.

$\therefore$              KE_f =qΔV

                            $\frac{P^{2}}{2m}=q\triangle V$

$\Rightarrow$                  $P= \sqrt{2mq\triangle V}$

 Since,  efficiency is 100% , number of eletrons emitted per second= number of photons incident per second.

    As, photon  is completely absorbed, force excerted

$=n(mV)=nP=n \sqrt{2mq\triangle V}$

= $\frac{200}{6.25\times 1.6\times10^{-19}}\times\sqrt{2(9\times10^{-31})\times1.6\times10^{-19}\times500}$

= 24