Question

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

• A. 1.8 eV
• B. 1.1 eV
• C. 0.8 eV
• D. 1.6 eV

Answer 1

Answer: (B)

Thinking Process The problem is based on frequency dependence of photoelectric emission. When incident light with certain frequency (greater than on the threshold frequency is focus 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 $r=\frac{mv}{eB}\Rightarrow \frac{r^2{e}^2{B}^2}{2}=\frac{m^2{v}^2}{2}$ $K{E}_{max}=\frac{(mv{)}^2}{2m}\Rightarrow \frac{r^2{e}^2{B}^2}{2m}=(KE{)}_{max}$ Work function of the metal (W), i.e. $W=hv-K{E}_{max}$ $1.89-\oint =\frac{r^2{e}^2{B}^2}{2m}\frac{1}{2}eV=\frac{r^{2}e{B}^2}{2m}eV$ [$hv\to 1.89eV,$ for the transition on from third to second orbit of H-atom] $=\frac{100\times 10^{-6}\times 1.6\times 10^{-19}\times 9\times 10^{-8}}{2\times 9.1\times 10^{-31}}$ $\oint =1.89-\frac{1.6\times 9}{2\times 9.1}$ $=1.89-0.79=1.1eV$