The radiation corresponding to 3→ 2 transition of hydrogen atom falls on a metal surface to produce photo electrons . These electrons are made toi enter a magnetic field of  $3\times 10^{-4}T$ . If the radius of the largest circular path followed by these electrons is 10.0mm. the work function of the metal is close to 

A) 1.8eV

B) 1.1eV

C) 0.8 eV

D) 1.6 eV


Option B


The problem is based on frequency dependence of photoelectric emission. When incident light with certain frequency (greater than on the threshold frequency is focused on a metal surface) then some electrons are emitted from the metal with substantial initial speed.

 When an electron moves in a circular path, then


$\Rightarrow $       $\frac{r^{2}e^{2}B^{2}}{2}=\frac{m^{2}v^{2}}{2}$


             $\Rightarrow$         $\frac{r^{2}e^{2}B^{2}}{2m}=KE_{max}$

 Work function  of the metal(W)

i.e,  $W=hv=KE_{max}$

  $1.89-\phi= \frac{r^{2}e^{2}B^{2}}{2m}\frac{1}{2}eV$

                             $= \frac{r^{2}e^{}B^{2}}{2m}eV$

[hv→1.89 eV, for the transition on from third to second orbit of H-atom]

=  $\frac{100\times10^{-6}\times1.6\times 10^{-19}\times9\times 10^{-8}}{2\times9.1\times10^{-31}}$


   =1.89-0.79=1.1 eV