Question

In Fig. electromagnetic radiations of wavelength 200nm are incident on a metallic plate A. the photoelectrons are accelerated by a potential difference of 10 V. These electrons strike another metal plate B from which electromagnetic radiations are emitted. The minimum wavelength of emitted photons is 100nm. If the work function of metal A is found to be `(xxx10^-1)ev`, then find the value of x. (Given `hc=1240eV` nm)

Answer 1

Incident energy of photons `=(hc)/(200)=6.2eV`
`6.2=K_1+W_0`,`K_1=6.2-W_0`
`K_2=K_1+10e=16.2-W_0`
`K_2=(hc)/(lamda)=12.4`,`W_0=3.8ev`, given` W_0=(x)/(10)`
`x=38`